Download Algebraic Problems and Exercises for High School (Sets, sets

Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
ALGEBRAIC PROBLEMS
AND EXERCISES
FOR HIGH SCHOOL
The Educational Publisher
Columbus, 2015
Algebraic problems and exercises for high school
Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
ALGEBRAIC PROBLEMS AND EXERCISES FOR HIGH SCHOOL
Sets, sets operations ■ Relations, functions ■ Aspects of combinatorics
1
Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
First imprint: Algebra în exerciții și probleme pentru liceu (in Romanian),
Cartier Publishing House, Kishinev, Moldova, 2000
Translation to English by Ana Maria Buzoianu
(AdSumus Cultural and Scientific Society)
The Educational Publisher
Zip Publishing
1313 Chesapeake Ave.
Columbus, Ohio 43212, USA
Email: [email protected]
AdSumus Cultural and Scientific Society
Editura Ferestre
13 Cantemir str.
410473 Oradea, Romania
Email: [email protected]
ISBN 978-1-59973-342-5
2
Algebraic problems and exercises for high school
Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
ALGEBRAIC PROBLEMS
AND EXERCISES
FOR HIGH SCHOOL
■ Sets, sets operations
■ Relations, functions
■ Aspects of combinatorics
The Educational Publisher
Columbus, 2015
3
Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
© Ion Goian, Raisa Grigor, Vasile Marin, Florentin Smarandache,
and the Publisher, 2015.
4
Algebraic problems and exercises for high school
Foreword
In this book, you will find algebraic exercises and problems,
grouped by chapters, intended for higher grades in high schools or
middle schools of general education. Its purpose is to facilitate
training in mathematics for students in all high school categories,
but can be equally helpful in a standalone work. The book can also
be used as an extracurricular source, as the reader shall find
enclosed important theorems and formulas, standard definitions
and notions that are not always included in school textbooks.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
Contents
Foreword / 5
Notations / 7
1. Sets. Operations with sets / 9
1.1. Definitions and notations / 9
1.2. Solved exercises / 16
1.3. Suggested exercises / 32
2. Relations, functions / 43
2.1. Definitions and notations / 43
2.2. Solved exercises / 59
2.3. Suggested exercises / 86
3. Elements of combinatorics / 101
3.1. Permutations. Arrangements. Combinations.
Newton’s binomial / 101
3.2. Solved problems / 105
3.3. Suggested exercises / 125
Answers / 137
References / 144
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Algebraic problems and exercises for high school
Notations
=
≠
∈
∉
⊆
⊇
∪
∩
∅
∨
∧
equality;
inequality;
belongs to;
doesn’t belong to;
subset;
superset;
union;
intersection;
empty set;
(or) disjunction;
(and) conjunction;
𝑑𝑒𝑓
𝑝⇔ 𝑞
ℕ = {0, 1, 2, 3, 4, … }
ℤ = {… , −2, −1, 0, 1, 2, … }
𝑚
ℚ = { 𝑛 | 𝑚, 𝑛 ∈ ℤ, 𝑛 ≠ 0}
p equivalent with 𝑞;
natural numbers
integer numbers set;
rational numbers set;
ℝ
real numbers set;
2
ℂ = {𝑎 + 𝑏𝑖|𝑎, 𝑏 ∈ 𝑅, 𝑖 = −1} complex numbers set;
𝐴+ = {𝑥 ∈ 𝐴|𝑥 > 0}, 𝐴 ∈ {ℤ, ℚ, ℝ};
𝐴− = {𝑦 ∈ 𝐴|𝑦 < 0}, 𝐴 ∈ {ℤ, ℚ, ℝ};
𝐴∗ = {𝑧 ∈ 𝐴|𝑧 ≠ 0} = 𝐴 ∖ {0}, 𝐴 ∈ {ℕ, ℤ, ℚ, ℝ, ℂ};
|𝑥|
absolute value of 𝑥 ∈ ℝ;
[𝑥]
integer part of x ∈ R;
{𝑥}
fractional part of 𝑥 ∈ ℝ,
0 ≤ {𝑥} < 1;
(𝑎, 𝑏)
pair with the first element 𝑎
and the second element 𝑏
(also called "ordered pair");
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
(𝑎, 𝑏, 𝑐)
triplet with corresponding elements
𝑎, 𝑏, 𝑐;
𝐴 × 𝐵 = {𝑎, 𝑏|𝑎 ∈ 𝐴, 𝑏 ∈ 𝐵}
Cartesian product of set
𝐴 and set 𝐵;
𝐴 × 𝐵 × 𝐶 = {𝑎, 𝑏, 𝑐|𝑎 ∈ 𝐴, 𝑏 ∈ 𝐵, 𝑐 ∈ 𝐶}
Cartesian product of sets 𝐴, 𝐵, 𝐶;
𝐸
universal set;
𝑃(𝐸) = {𝑋|𝑋 ⊆ 𝐸}
set of parts (subsets) of set 𝐸;
𝑑𝑒𝑓
𝐴 = 𝐵 ⇔ (∀) 𝑥 ∈ 𝐸(𝑥 ∈ 𝐴 ⇔ 𝑥 ∈ 𝐵)
equality of sets 𝐴 and 𝐵;
𝑑𝑒𝑓
𝐴 ⊆ 𝐵 ⇔ (∀) 𝑥 ∈ 𝐸(𝑥 ∈ 𝐴 ⇔ 𝑥 ∈ 𝐵)
𝐴 is included in 𝐵;
𝐴⋃𝐵 = {𝑥 ∈ 𝐸|𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵} union of sets 𝐴 and 𝐵;
𝐴⋂𝐵 = {𝑥 ∈ 𝐸|𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵} intersection of sets 𝐴 and 𝐵;
𝐴 ∖ 𝐵 = {𝑥 ∈ 𝐸|𝑥 ∈ 𝐴 ∧ 𝑥 ∉ 𝐵} difference between sets 𝐴 and 𝐵;
𝐴∆𝐵 = (𝐴 ∖ 𝐵) ∪ (𝐵 ∖ 𝐴)
symmetrical difference;
̅
(𝐴)
∁𝜀
=𝐴=𝐸∖𝐴
complement of set 𝐴 relative to 𝐸;
𝛼 ⊆𝐴×𝐵
relation 𝛼 defined on sets 𝐴 and 𝐵;
𝑓: 𝐴 → 𝐵
function (application) defined on 𝐴
with values in 𝐵;
𝐷(𝑓)
domain of definition of function 𝑓:
𝐸(𝑓)
domain of values of function 𝑓.
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Algebraic problems and exercises for high school
1. Sets. Operations with sets
1.1. Definitions and notations
It is difficult to give an account of the axiomatic theory of sets at
an elementary level, which is why, intuitively, we shall define a set as a
collection of objects, named elements or points of the set. A set is
considered defined if its elements are given or if a property shared by
all of its elements is given, a property that distinguishes them from the
elements of another set. Henceforth, we shall assign capital letters to
designate sets: 𝐴, 𝐵, 𝐶, … , 𝑋, 𝑌, 𝑍, and small letters for elements in sets:
𝑎, 𝑏, 𝑐, … , 𝑥, 𝑦, 𝑧 etc.
If 𝑎 is an element of the set 𝐴, we will write 𝑎 ∈ 𝐴 and we will
read "𝑎 belongs to 𝐴" or "𝑎 is an element of 𝐴". To express that 𝑎 is not
an element of the set 𝐴, we will write 𝑎 ∉ 𝐴 and we will read "𝑎 does
not belong to 𝐴".
Among sets, we allow the existence of a particular set, noted as
∅, called the empty set and containing no element.
The set that contains a sole element will be noted with {𝑎}.
More generally, the set that doesn’t contain other elements except the
elements 𝑎1 , 𝑎2 , … , 𝑎𝑛 will be noted by {𝑎1 , 𝑎2 , … , 𝑎𝑛 }.
If 𝐴 is a set, and all of its elements have the quality 𝑃, then we
will write 𝐴 = {𝑥|𝑥 verifies 𝑃} or 𝐴 = {𝑥|𝑃(𝑥)} and we will read: "𝐴
consists of only those elements that display the property 𝑃 (for which
the predicate 𝑃( 𝑥) is true)."
We shall use the following notations:
ℕ = {0, 1, 2, 3, … } – the natural numbers set;
ℕ∗ = {0, 1, 2, 3, … } – the natural numbers set without zero;
ℤ = {… , −2, −1, 0, 1, 2, … } – the integer numbers set;
ℤ = {±1, ±2, ±3 … } – the integer numbers set without zero;
𝑚
ℚ = { | 𝑚, 𝑛 ∈ ℤ, 𝑛 ∈ ℕ∗ } – the rational numbers set;
𝑛
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
ℚ∗ = the rational numbers set without zero;
ℝ = the real numbers set;
ℝ∗ = the real numbers set without zero;
ℝ+ = {𝑥 ∈ ℝ|𝑥 ≥ 0}; ℝ∗+ = {𝑥 ∈ ℝ|𝑥 > 0};
ℂ = {𝑎 + 𝑏𝑖|𝑎, 𝑏 ∈ 𝑅, 𝑖 2 = −1} = the complex numbers set;
ℂ∗ = the complex numbers set without zero;
𝑚 ∈ {1, 2, … , 𝑛} ⟺ 𝑚 = ̅̅̅̅̅
1, 𝑛;
𝐷(𝑎) = {𝑐 ∈ ℤ∗ |𝑎 ⋮ 𝑐} = the set of all integer divisors of number
𝑎 ∈ ℤ;
𝑛(𝐴) = |𝐴| = the number of the elements of finite set A.
Note. We will consider that the reader is accustomed to the
symbols of Logic: conjunction ∧ (…and…), disjunction ∨ (…or…),
implication ⟹, existential quantification (∃) and universal quantification
(∀).
Let 𝐴 and 𝐵 be two sets. If all the elements of the set 𝐴 are also
elements of the set 𝐵, we then say that 𝑨 is included in 𝑩, or that 𝑨 is
a part of 𝑩, or that 𝑨 is a subset of the set 𝑩 and we write 𝐴 ⊆ 𝐵. So
𝐴 ⊆ 𝐵 ⇔ (∀)𝑥(𝑥 ∈ 𝐴 ⟹ 𝑥 ∈ 𝐵).
The properties of inclusion
a. (∀) 𝐴, 𝐴 ⊆ 𝐴 (reflexivity);
b. (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐶) ⟹ 𝐴 ⊆ 𝐶 (transitivity);
c. (∀)𝐴, ∅ ⊆ 𝐴.
If 𝐴 is not part of the set 𝐵, then we write 𝐴 ⊈ 𝐵, that is, 𝐴 ⊈
𝐵 ⇔ (∃)𝑥(𝑥 ∈ 𝐴 ∧ 𝑥 ∉ 𝐵).
We will say that the set 𝐴 is equal to the set 𝐵, in short 𝐴 = 𝐵,
if they have exactly the same elements, that is
𝐴 = 𝐵 ⟺ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴).
The properties of equality
Irrespective of what the sets 𝐴, 𝐵 and 𝐶 may be, we have:
a. 𝐴 = 𝐴 (reflexivity);
b. (𝐴 = 𝐵) ⟹ (𝐵 = 𝐴) (symmetry);
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Algebraic problems and exercises for high school
c. (𝐴 = 𝐵 ∧ 𝐵 = 𝐶) ⟹ (𝐴 = 𝐶) (transitivity).
With 𝑃(𝐴) we will note the set of all parts of set 𝐀, in short
𝑋 ∈ 𝑃(𝐴) ⟺ 𝑋 ⊆ 𝐴.
Obviously, ∅, 𝐴 ∈ 𝑃(𝐴).
The universal set, the set that contains all the sets examined
further, which has the containing elements’ nature one and the same,
will be denoted by 𝐸.
Operations with sets
Let 𝐴 and 𝐵 be two sets, 𝐴, 𝐵 ∈ 𝑃(𝐸).
1. Intersection.
𝐴 ∩ 𝐵 = {𝑥 ∈ 𝐸|𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵},
i.e.
𝑥 ∈ 𝐴 ∩ 𝐵 ⟺ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵),
𝑥 ∉ 𝐴 ∩ 𝐵 ⟺ (𝑥 ∉ 𝐴 ∧ 𝑥 ∉ 𝐵),
2. Union.
𝐴 ∪ 𝐵 = {𝑥 ∈ 𝐸|𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵},
(1)
(1’)
𝑥 ∈ 𝐴 ∪ 𝐵 ⟺ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵),
𝑥 ∉ 𝐴 ∪ 𝐵 ⟺ (𝑥 ∉ 𝐴 ∨ 𝑥 ∉ 𝐵),
3. Difference.
𝐴 ∖ 𝐵 = {𝑥 ∈ 𝐸|𝑥 ∈ 𝐴 ∧ 𝑥 ∉ 𝐵},
(2)
(2’)
i.e.
i.e.
𝑥 ∈ 𝐴 ∖ 𝐵 ⟺ (𝑥 ∈ 𝐴 ∧ 𝑥 ∉ 𝐵),
(3)
𝑥 ∉ 𝐴 ∖ 𝐵 ⟺ (𝑥 ∉ 𝐴 ∨ 𝑥 ∈ 𝐵),
(3’)
4. The complement of a set. Let 𝐴 ∈ 𝑃(𝐸). The difference 𝐸 ∖
𝐴 is a subset of 𝐸, denoted 𝐶𝐸 (𝐴) and called “the complement of 𝐴
relative to 𝐸”, that is
𝐶𝐸 (𝐴) = 𝐸 ∖ 𝐴 = {𝑥 ∈ 𝐸|𝑥 ∉ 𝐴}.
In other words,
𝑥 ∈ 𝐶𝐸 (𝐴) ⟺ 𝑥 ∉ 𝐴,
(4)
𝑥 ∉ 𝐶𝐸 (𝐴) ⟺ 𝑥 ∈ 𝐴.
(4’)
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
Properties of operations with sets
𝐴 ∩ 𝐴 = 𝐴, 𝐴 ∪ 𝐴 = 𝐴 (idempotent laws)
𝐴 ∩ 𝐵 = 𝐵 ∩ 𝐴, 𝐴 ∪ 𝐵 = 𝐵 ∪ 𝐴 (commutative laws)
(𝐴 ∩ 𝐵) ∩ 𝐶 = 𝐴 ∩ (𝐵 ∩ 𝐶);
(𝐴 ∪ 𝐵) ∪ 𝐶 = 𝐴 ∪ (𝐵 ∪ 𝐶) (associativity laws)
𝐴 ∪ (𝐵 ∩ 𝐶) = (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶);
𝐴 ∩ (𝐵 ∪ 𝐶) = (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶) (distributive laws)
𝐴 ∪ (𝐴 ∩ 𝐵) = 𝐴;
𝐴 ∩ (𝐴 ∪ 𝐵) = 𝐴 (absorption laws)
𝐶𝐸 (𝐴 ∪ 𝐵) = 𝐶𝐸 (𝐴) ∩ 𝐶𝐸 (𝐵);
𝐶𝐸 (𝐴 ∩ 𝐵) = 𝐶𝐸 (𝐴) ∪ 𝐶𝐸 (𝐵) (Morgan’s laws)
Two "privileged" sets of 𝐸 are ∅ and 𝐸.
For any 𝐴 ∈ 𝑃(𝐸), we have:
∅ ⊆ 𝐴 ⊆ 𝐸,
𝐴 ∪ ∅ = 𝐴,
𝐴 ∩ ∅ = ∅,
𝐶𝐸 (∅) = 𝐸,
𝐴 ∪ 𝐸 = 𝐸,
𝐴 ∩ 𝐸 = 𝐴,
𝐶𝐸 (𝐸) = ∅,
𝐴 ∪ 𝐶𝐸 (𝐴) = 𝐸,
𝐴 ∩ 𝐶𝐸 (𝐴) = ∅,
𝐶𝐸 (𝐶𝐸 (𝐴)) = 𝐴 (principle of reciprocity).
Subsequently, we will use the notation 𝐶𝐸 (𝐴) = 𝐴̅.
5. Symmetric difference.
𝐴∆𝐵 = (𝐴 ∖ 𝐵) ∪ (𝐵 ∖ 𝐴).
Properties. Irrespective of what the sets 𝐴, 𝐵 and 𝐶 are, we
have:
a. 𝐴∆𝐴 = ∅;
b. 𝐴∆𝐵 = 𝐵∆𝐴 (commutativity);
c. 𝐴∆∅ = ∅∆𝐴 = 𝐴;
d. 𝐴∆ (𝐴∆𝐵) = 𝐵;
e. (𝐴∆𝐵)∆𝐶 = 𝐴∆(𝐵∆𝐶) (associativity);
f. 𝐴 ∩ (𝐵∆𝐶) = (𝐴 ∩ 𝐵)∆(𝐴 ∩ 𝐶);
g. 𝐴∆𝐵 = (𝐴 ∪ 𝐵) ∖ (𝐴 ∩ 𝐵).
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Algebraic problems and exercises for high school
6. Cartesian product. Let 𝑥 and 𝑦 be two objects. The set
{{𝑥}, {𝑥, 𝑦}}, whose elements are the sets {𝑥} and {𝑥, 𝑦}, is called an
ordered pair (or an ordered couple) with the first component 𝑥 and
the second component 𝑦 and is denoted as (𝑥, 𝑦). Having three objects
𝑥, 𝑦 and 𝑧, we write (𝑥, 𝑦, 𝑧) = ((𝑥, 𝑦), 𝑧) and we name it an ordered
triplet.
Generally, having 𝑛 objects 𝑥1 , 𝑥2 , … , 𝑥𝑛 we denote
(𝑥1 , 𝑥2 , … , 𝑥𝑛 ) = (… ((𝑥1 , 𝑥2 )𝑥3 ) … 𝑥𝑛 )
and we name it an ordered system of 𝑛 elements (or a cortege of
length 𝑛).
We have
(𝑥1 , 𝑥2 , … , 𝑥𝑛 ) = (𝑦1 , 𝑦2 , … , 𝑦𝑛 )
⟺ (𝑥1 = 𝑦1 ∧ 𝑥2 = 𝑦2 ∧ … ∧ 𝑥𝑛 = 𝑦𝑛 ).
Let 𝐴, 𝐵 ∈ 𝑃(𝐸). The set
𝐴 × 𝐵 = {(𝑎, 𝑏)|𝑎 ∈ 𝐴 ∧ 𝑏 ∈ 𝐵}
is called a Cartesian product of sets 𝐴 and 𝐵. Obviously, we can define
𝐴 × 𝐵 × 𝐶 = {(𝑥, 𝑦, 𝑧)|𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝑧 ∈ 𝐶}.
More generally, the Cartesian product of the sets 𝐴1 , 𝐴2 , … , 𝐴𝑛
𝐴1 × 𝐴2 × … × 𝐴𝑛 = {(𝑥1 , 𝑥2 , … , 𝑥𝑛 )|𝑥𝑖 ∈ 𝐴𝑖 , 𝑖 = ̅̅̅̅̅
1, 𝑛}.
For 𝐴 = 𝐵 = 𝐶 = 𝐴1 = 𝐴2 = . . . = 𝐴𝑛 , we have
𝐴 × 𝐴 ≝ 𝐴2 , 𝐴 × 𝐴 × 𝐴 ≝ 𝐴3 , ⏟
𝐴 × 𝐴 × … × 𝐴 ≝ 𝐴𝑛 .
𝑛 ori
For example, ℝ3 = {(𝑥, 𝑦, 𝑧)|𝑥, 𝑦, 𝑧 ∈ ℝ}. The subset
Δ = {(𝑎, 𝑎)|𝑎 ∈ 𝐴} ⊆ 𝐴2
is called the diagonal of set 𝐴2.
Examples.
1. Let 𝐴 = {1,2} and 𝐵 = {1, 2, 3}. Then
𝐴 × 𝐵 = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3)}
and
𝐵 × 𝐴 = {(1,1), (1,2), (2,1), (2,2), (3,1), (3,2)}.
We notice that 𝐴 × 𝐵 ≠ 𝐵 × 𝐴.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
2. The Cartesian product ℝ2 = ℝ × ℝ can be geometrically
represented as the set of all the points from a plane to wich a
rectangular systems of coordinates 𝑥𝑂𝑦 has been assigned,
associating to each element (𝑥, 𝑦) ∈ ℝ2 a point 𝑃(𝑥, 𝑦) from the plane
of the abscissa 𝑥 and the ordinate 𝑦.
Let 𝐴 = [2; 3] and 𝐵 = [1; 5], (𝐴, 𝐵 ⊆ ℝ). Then 𝐴 × 𝐵 can be
geometrically represented as the hatched rectangle 𝐾𝐿𝑀𝑁 (see Figure
1.1), where 𝐾(2,1), 𝐿(2,5), 𝑀(3,5), 𝑁(3,1).
Figure 1.1.
The following properties are easily verifiable:
a. (𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷) ⟹ 𝐴 × 𝐵 ⊆ 𝐶 × 𝐷;
b. 𝐴 × (𝐵 ∪ 𝐶) = (𝐴 × 𝐵) ∪ (𝐴 × 𝐶),
𝐴 × (𝐵 ∩ 𝐶) = (𝐴 × 𝐵) ∩ (𝐴 × 𝐶);
c. 𝐴 × 𝐵 = ∅ ⟺ (𝐴 = ∅ ∨ 𝐵 = ∅),
𝐴 × 𝐵 ≠ ∅ ⟺ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅).
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Algebraic problems and exercises for high school
7. The intersection and the union of a family of sets. A family of
sets is a set {𝐴𝑖 |𝑖 ∈ 𝐼} = {𝐴𝑖 }𝑖∈𝐼 whose elements are the sets 𝐴𝑖 , 𝑖 ∈
𝐼, 𝐴𝑖 ∈ 𝑃(𝐸). We say that 𝐴𝑖 , 𝑖 ∈ 𝐼, 𝐴𝑖 ∈ 𝑃(𝐸) is a family of sets
indexed with the set 𝐼.
Let there be a family of sets {𝐴𝑖 |𝑖 ∈ 𝐼}. Its union (or the union of
the sets 𝐴𝑖 , 𝑖 ∈ 𝐼) is the set
.𝑖 ∈𝐼
The intersection of the given family (or the intersection of the
sets 𝐴𝑖 , 𝑖 ∈ 𝐼) is the set
In the case of 𝐼 = {1,2, … , 𝑛}, we write
8. Euler-Wenn Diagrams. We call Euler Diagrams (in USA –
Wenn’s diagrams) the figures that are used to interpret sets (circles,
squares, rectangles etc.) and visually illustrate some properties of
operations with sets. We will use the Euler circles.
Example. Using the Euler diagrams, prove Morgan’s law.
Solution.
In fig. 1.2.a, the hatched part is 𝐴 ∩ 𝐵; the uncrossed one
(except 𝐴 ∩ 𝐵) represents 𝐶𝐸 (𝐴 ∩ 𝐵).
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
Figure 2.2.a
Figure 3.2.b
In fig. 1.2b the side of the square hatched with “ \ \ \ \ ” is equal
to 𝐶𝐸 (𝐴) and the one hatched with “ / / / / ” is equal to 𝐶𝐸 (𝐵). All the
hatched side forms 𝐶𝐸 (𝐴) ∪ 𝐶𝐸 (𝐵) (the uncrossed side is exactly 𝐴 ∩
𝐵).
From these two figures it can be seen that 𝐶𝐸 (𝐴 ∩ 𝐵) (the
uncrossed side of the square in fig. 1.2.a coincides with 𝐶𝐸 (𝐴) ∪
𝐶𝐸 (𝐵) (the hatched side of fig. 1.2.b), meaning
𝐶𝐸 (𝐴 ∩ 𝐵) = 𝐶𝐸 (𝐴) ∪ 𝐶𝐸 (𝐵).
1.2. Solved exercises
1. For any two sets 𝐴 and 𝐵, we have
𝐴 ∩ 𝐵 = 𝐴 \ (𝐴\𝐵)
Solution. Using the definitions from the operations with sets, we
gradually obtain:
From this succession of equivalences it follows that:
which proves the required equality.
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Algebraic problems and exercises for high school
Remark. The equality can also be proven using Euler’s diagrams.
So 𝐴 ∩ 𝐵 = 𝐴\(𝐴\𝐵).
2. Whatever 𝐴, 𝐵 ⊆ 𝐸 are, the following equality takes place:
Solution. The analytical method. Using the definitions from
the operations with sets, we obtain:
This succession of equivalences proves that the equality from
the enunciation is true.
The graphical method. Using Euler’s circles, we have
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
Figure 4.3.a
Figure 5.3.b
In fig. 1.3.a, we have (𝐴 ∩ 𝐵) ∪ ( 𝐴 ∩ 𝐵), which represents the
hatched side of the square. From fig. 1.3.b it can be seen that
(𝐴 ∩ 𝐵) ∪ ( 𝐴 ∩ 𝐵) = (𝐴 ∪ 𝐵) ∩ (𝐴 ∩ 𝐵).
3. For every two sets 𝐴, 𝐵 ⊆ 𝐸, the equivalence stands true:
Solution. Let 𝐴\𝐵 = 𝐵\𝐴. We assume that 𝐴 ≠ 𝐵. Then there
is 𝑎 ∈ 𝐴 with 𝑎 ∉ 𝐵 or 𝑏 ∈ 𝐵 with 𝑏 ∉ 𝐴.
In the first case we obtain 𝑎 ∈ 𝐴\𝐵 and 𝑎 ∉ 𝐵\𝐴 , which
contradicts the equality 𝐴\𝐵 = 𝐵\𝐴. In the second case, we obtain
the same contradiction.
So, if 𝐴\𝐵 = 𝐵\𝐴 ⟹ 𝐴 = 𝐵.
Reciprocally, obviously.
4. Sets 𝐴 = {1,2,3,4,5,6,7,8,9,10}, 𝐵 = {2,4,6,8,10} and 𝐶 =
{3,6,9} are given. Verify the following equalities:
a) 𝐴\(𝐵 ∪ 𝐶) = (𝐴\𝐵) ∩ (𝐴\𝐶);
b) 𝐴\(𝐵 ∩ 𝐶) = (𝐴\𝐵) ∪ (𝐴\𝐶).
Solution. a) We have 𝐵 ∪ 𝐶 = {2,3,4,6,8,9,10} , 𝐴\(𝐵 ∪ 𝐶) =
{1,5,7}, 𝐴\𝐵 = {1,3,5,7,9}, 𝐴\𝐶 = {1,2,4,5,7,8,10}, (𝐴\𝐵) ∩ (𝐴\𝐶) =
{1,5,7} = 𝐴\(𝐵 ∪ 𝐶).
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Algebraic problems and exercises for high school
b) For the second equality, we have
𝐵 ∩ 𝐶 = {6} , 𝐴\(𝐵 ∩ 𝐶) = {1,2,3,4,5,6,7,8,9,10} , (𝐴\𝐵) ∪
(𝐴\𝐶) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} = 𝐴\(𝐵 ∩ 𝐶).
5. Determine the sets A and B that simultaneously satisfy the
following the conditions:
1) 𝐴 ∪ 𝐵 = {1,2,3,4,5};
2) 𝐴 ∩ 𝐵 = {3,4,5};
3)1 ∉ 𝐴\𝐵;
4) 2 ∉ 𝐵\𝐴.
Solution. From 1) and 2) it follows that {3,4,5} ⊆ 𝐴 ⊆ 𝐴 ∪ 𝐵 and
{3,4,5} ⊆ 𝐵 ⊆ 𝐴 ∪ 𝐵. From 3) it follows that 1 ∉ 𝐴 or 1 ∈ 𝐵. If 1 ∉ 𝐴,
then from 𝐴 ∪ 𝐵 = {1,2,3,4,5} it follows that 1 ∈ 𝐵. But, if 1 ∈ 𝐵,
then 1 ∉ 𝐴, because, on the contrary it would follow that 1 ∈ 𝐴 ∩ 𝐵 =
{3,4,5}. So 1 ∈ 𝐵 and 1 ∉ 𝐴 remain. Similarly, from 4) it follows that
2 ∉ 𝐵 and so 2 ∈ 𝐴 . In other words, {3,4, 5} ⊆ 𝐴 ⊆ {2,3,4,5} and
{3, 4, 5} ⊆ 𝐵 ⊆ {1,3,4,5} with 2 ∈ 𝐴 ∪ 𝐵 , 1 ∈ 𝐴 ∪ 𝐵 and that is why
𝐴 = {2,3,4,5}, and 𝐵 = {1,3,4,5}.
Answer: 𝐴 = {2,3,4,5}, 𝐵 = {1,3,4,5}.
6. The sets 𝐴 = {11𝑘 + 8│𝑘 ∈ ℤ}, 𝐵 = {4𝑚│𝑚 ∈ ℤ} and 𝐶 =
{11(4𝑛 + 1) − 3│𝑛 ∈ ℤ}, are given. Prove that 𝐴 ∩ 𝐵 = 𝐶.
Solution. To obtain the required equality, we will prove the truth
of the equivalence:
𝑥 ∈ 𝐴 ∩ 𝐵 ⟺ 𝑥 ∈ 𝐶.
Let 𝑥 ∈ 𝐴 ∩ 𝐵 . Then 𝑥 ∈ 𝐴 and 𝑥 ∈ 𝐵 and that is why two
integer numbers 𝑘, 𝑚 ∈ ℤ , exist, so that 𝑥 = 11𝑘 + 8 = 4𝑚 ⟺
11𝑘 = 4(𝑚 − 2). In this equality, the right member is divisible by 4,
and 11, 4 are prime. So, from 11𝑘 ⋮ 4 it follows that 𝑘 ⋮ 4, giving 𝑘 =
4𝑡 for one 𝑡 ∈ ℤ. Then
𝑥 = 11𝑘 + 8 = 11 ∙ 4𝑡 + 8 = 11 ∙ 4𝑡 + 11 − 3 = 11(4𝑡 + 1) − 3
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
which implies 𝑥 ∈ 𝐶, in other words, we have proven the implication
𝑥 ∈ 𝐴 ∩ 𝐵 ⇒ 𝑒 ∈ 𝐶.
(1)
Similarly, let 𝑦 ∈ 𝐶. Then 𝑠 ∈ ℤ exists with
𝑦 = 11(4𝑠 + 1) − 3 = 11 ∙ 4𝑠 + 11 − 3 = 11 ∙ 4𝑠 + 8 = 4(11𝑠 + 2)
Taking 4𝑠 = 𝑢 ∈ ℤ and 11𝑠 + 2 = 𝑣 ∈ ℤ, we obtain
𝑦 = 11𝑢 + 8 = 4𝑣 ∈ 𝐴 ∩ 𝐵,
which proves the truth of the implication
𝑦 ∈ 𝐶 ⇒ 𝑦 ∈ 𝐴 ∩ 𝐵.
(2)
From (1) and (2) the required equality follows.
7. The following sets are given
and 𝐵 = 𝐴 ∩ ℕ.
Prove that:
a) all 𝑋 sets with 𝐵 ∪ 𝑋 = {3,4,5,6,7,8,9};
b) all 𝑌 = {𝑦 ∈ ℤ│𝑦 2 ∈ 𝐵 ∪ 𝑋}, so that 𝐵 ∩ 𝑌 = {3}.
Solution. We determine the set 𝐴:
Then 𝐵 = (3; 6) ∩ 𝐼𝑁 = {3,4,5}.
a) All possible subsets of 𝐵 are
∅, {3}{4}{5}{3,4}{3,5}{4,5}{3,4,5} = 𝐵
The required 𝑋 sets are such that 𝑋 ∪ 𝐵 = {3,4,5,6,7,8,9} and
will thus be like 𝑋 = 𝐶 ∪ {6,7,8,9}, where 𝐶 ∈ 𝑃(𝐵), namely, the sets
required at point a) are:
𝑋1 {6,7,8,9} , 𝑋2 {3, 6, 7,8, 9} , 𝑋3 {4, 6, 7,8, 9} , 𝑋4 {5,6, 7,8,9} ,
{3,4,6,7,8,9},
𝑋5
𝑋6 {3,5,6,7,8,9}, 𝑋7 {4,5,6, 7,8,9}, 𝑋8 {3,4,5,6, 7,8,9}..
b) Because 𝑦 ∈ ℤ, then
and vice versa. Considering
{3,4,5,6,7,8,9},
that 𝑦 ∈ 𝐵 ∪ 𝑋 =
we obtain 𝑦 2 ∈ {4,9}, namely 𝑦 2 ∈
{−3, −2,2,3} = 𝑀. The parts of the set 𝑀 are:
2
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Algebraic problems and exercises for high school
∅ , {−3} , {−2} , {2} , {3} , {−3, −2} , {−3,2} , {−3,3} , {−2,2} ,
{−2,3}, {2,3}, {−3, −2,2}, {−3, −2,3}, {−3,2,3}, {−2,2,3}, 𝑀.
From the condition 𝐵 ∩ 𝑌 = {3} it follows that 𝑌 is one of the
sets
𝑌1 = {3}, 𝑌2 = {−3,3}, 𝑌3 = {−2,3}, 𝑌4 = {2,3}, 𝑌5 = {−3, −2,3},
{−3,2,3},
𝑌6 =
𝑌7 = {−2,2,3}, 𝑌8 = 𝑀 = {−3, −2,2,3}.
Answer: a) 𝑋 ∈ {𝑋1 , 𝑋 2 , 𝑋 3 , 𝑋 4 , 𝑋 5 , 𝑋 6 , 𝑋 7 , 𝑋 8 };
b) 𝑌 ∈ {𝑌1 , 𝑌 2 , 𝑌 3 , 𝑌 4 , 𝑌 5 , 𝑌 6 , 𝑌 7 , 𝑌 8 }.
8. Determine 𝐴, 𝐵, 𝐶 ⊆ 𝑇 and 𝐴 ∆ 𝐵, if
𝑇 = {1,2,3,4,5,6} , 𝐴 ∆ 𝐶 = {1,2} , 𝐵 ∆ 𝐶 = {5,6} , 𝐴 ∩ 𝐶 = 𝐵 ∩
𝐶 = {3,4}.
Solution. From 𝐴 ∩ 𝐶 = 𝐵 ∩ 𝐶 = {3,4}, it follows that {3,4} ⊆
𝐴 ∩ 𝐵 ∩ 𝐶.
We know that
𝐴 ∆ 𝐶 = (𝐴 ∖ 𝐶) ∪ (𝐶 ∖ 𝐴) = (𝐴 ∪ 𝐶) ∖ (𝐴 ∩ 𝐶),
𝐴 ∆ 𝐶 = (𝐵 ∖ 𝐶) ∪ (𝐶 ∖ 𝐵) = (𝐵 ∪ 𝐶) ∖ (𝐵 ∩ 𝐶).
Then:
1 ∈ 𝐴 ∆ 𝐶 ⇔ (1 ∈ 𝐴 ∪ 𝐶 ∧ 1 ∉ 𝐴 ∩ 𝐶) ⟺ ((1 ∈ 𝐴 ∨ 1 ∈ 𝐶) ∧
1 ∉ 𝐴 ∩ 𝐶).
The following cases are possible:
a) 1 ∉ 𝐴 and1 ∈ 𝐶;
b) 1 ∈ 𝐴 and 1 ∉ 𝐶
(case no. 3, 1 ∈ 𝐴 and 1 ∈ 𝐶 ⇒ 1 ∈ 𝐴 ∩ 𝐶 = {3,4}, is impossible).
In the first case 1 ∉ 𝐴 and 1 ∈ 𝐶, from 𝐵 ∆ 𝐶 = {5,6} it follows
that 1 ∈ 𝐵 , because otherwise 1 ∉ 𝐵 and 1 ∈ 𝐶 ⇒ 1 ∈ 𝐶 ∖ 𝐵 ⊆
𝐵∆ 𝐶 = {5,6}. So, in this case we have 1 ∈ 𝐵 ∩ 𝐶 = {3,4}which is
impossible, so it follows that 1 ∈ 𝐴, 1 ∉ 𝐶. Similarly, we obtain 2 ∈
𝐴 and 2 ∉ 𝐵 and 2 ∉ 𝐶 , 5 ∈ 𝐵 and 5 ∉ 𝐴 ,5 ∉ 𝐶 and 6 ∈ 𝐵 , 6 ∉ 𝐴, 6 ∉
𝐴.
In other words, we have obtained:
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
𝐴 = {1,2,3,4}, 𝐵 = {3,4,5,6}, 𝐶 = {3,4} and 𝐴 ∆ 𝐵 = {1,2,5,6}.
Answer: 𝐴 = {1,2,3,4}, 𝐵 = {3,4,5,6}, 𝐶 = {3,4} and
𝐴 ∆ 𝐵 = {1,2,5,6}.
9. The sets 𝐴 = {1,2}, 𝐵 = {2,3} are given. Determine the following sets:
a) 𝐴 × 𝐵;
b) 𝐵 × 𝐴;
c) 𝐴2;
2
d) 𝐵 ;
e) (𝐴 × 𝐵) ∩ (𝐵 × 𝐴); f) (𝐴 ∪ 𝐵) × 𝐵;
g) (𝐴 × 𝐵) ∪ (𝐵 × 𝐵).
Solution. Using the definition for the Cartesian product with
two sets, we obtain:
a) 𝐴 × 𝐵 = {(1,2), (1,3), (2,2), (2,3)};
b) 𝐵 × 𝐴 = {(2,1), (2,2), (3,1), (3,2)};
c) 𝐴2 = {(1,1), (1,2), (2,1), (2,2)};
d) 𝐵2 = {(2,2), (2,3), (3,2), (3,3)};
e) (𝐴 × 𝐵) ∩ (𝐵 × 𝐴) = {(2,2)};
f)𝐴 ∪ 𝐵 = {1,2,3}; (𝐴 ∪ 𝐵) × 𝐵 = {(1,2), (1,3), (2,2)
(2,3), (3,2), (3,3)};
g) (𝐴 × 𝐵) ∪ (𝐵 × 𝐵) = {(1,2), (1,3), (2,2), (2,3),
(3,2), (3,3)} = (𝐴 ∪ 𝐵) × 𝐵.
10. The sets 𝐴 = {1,2, 𝑥}, 𝐵 = {3,4, 𝑦} are given. Determine 𝑥
and 𝑦, knowing that {1,3} × {2,4} ⊆ 𝐴 × 𝐵.
Solution. We form the sets 𝐴 × 𝐵 and {1,3} × {2,4} :
Because {1,3} × {2,4} ⊆ 𝐴 × 𝐵, we obtain
For 𝑥 = 3 and 𝑦 = 2, we have (3,2) ∈ 𝐴 × 𝐵.
Answer: 𝑥 = 3, 𝑦 = 2.
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Algebraic problems and exercises for high school
11. If 𝐴 ⊇ 𝐵, then
Prove.
Solution. 𝐵 ⊇ 𝐴 ⇒ (𝐴 ∖ 𝐵) ∪ 𝐵 = 𝐴 and that is why
(we have used the equality (𝐴 ∪ 𝐵) × 𝐶 = (𝐴 × 𝐶) ∪ (𝐵 × 𝐶)).
12. How many elements does the following set have:
Solution. The set 𝐴 has as many elements as the fraction
(𝑛2 + 1) ∕ (2𝑛2 + 𝑛 + 1) has different values, when 𝑛 takes the values
1,2,3, … , 1000. We choose the values of 𝑛 for which the fraction takes
equal values.
Let 𝑚, 𝑙 ∈ ℕ∗ , 𝑚 < 1 with
Then
But 𝑚 ∈ ℕ∗ and consequently
For 𝑙 = 2, we obtain 𝑚 = 3, and for 𝑙 = 3, we have 𝑚 = 2.
Considering that 𝑚 < 𝑙, we obtain 𝑚 = 2 and 𝑙 = 3. So, only for 𝑛 =
2 and 𝑛 = 3, the same element of the set 𝐴: 𝑥 = 5 ∕ 11 is obtained.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
Answer: the set 𝐴 has 999 elements, namely 𝑛(𝐴) = 999.
13. Determine the integer numbers 𝑥, 𝑦 that make the following
statement true (𝑥 − 1) ⋅ (𝑦 − 3) = 13.
Solution. We write
As 13 is a prime number, and 𝑥, 𝑦 ∈ ℤ, the following cases are
possible:
meaning the proposition 𝑃(𝑥) is true only in these situations:
14. Determine the set
Solution. Because:
it follows that that equality
is possible, if and only if we simultaneously have:
Then:
a) if at least two of the numbers 𝑎, 𝑏 and 𝑐 are different, we
cannot have the equality:
namely in this case 𝐴 = ∅;
b) if 𝑎 = 𝑏 = 𝑐, then 𝑥 = −𝑎 and 𝐴 = {−𝑎}.
Answer: 1) for 𝑎 = 𝑏 = 𝑐, we have 𝐴 = {−2}; 2) if at least two of
the numbers 𝑎, 𝑏 and 𝑐 are different, then 𝐴 = ∅.
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Algebraic problems and exercises for high school
15. Determine the set
Solution. Possible situations:
𝑥 + 2 ≤ 4 − 𝑥/3 or 𝑥 + 2 > 4 − 𝑥/3.
We examine each case:
In this case we have:
All integer numbers 𝑥 for which −1 ≤ 𝑥 ≤ 3 ∕ 2 applies, are:
−1,0,1.
Then
All integer numbers 𝑥 for which 3 ∕ 2 < x ≤ 9 applies, are:
2,3,4,5,6,7,8,9.
We thus obtain:
Answer: 𝐴 = {−1,0,1,2,3,4,5,6,7,8,9}.
16. Determine the values of the real parameter 𝑚 for which the
set
has:
a) one element;
b) two elements;
c) is null.
Solution. The set 𝐴 coincides with the set of the solutions of
the square equation
and the key to the problem is reduced to determining the values of the
parameter 𝑚 ∈ ℝ for which the equation has one solution, two
different solutions or no solution.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
a) The equation (1) has one (two equal) solution, if and only if
𝐷 = 0 for 𝑎 = 𝑚 − 1 ≠ 0 or if 𝑎 = 𝑚 − 1 = 0.
We examine these cases:
2) For 𝑚 = 1, the equation (1) becomes −5𝑥 + 15 = 0 ⇔ 𝑥 = 3. So,
the set 𝐴 consists of one element for
b) The equation (1) has two different roots, if and only if 𝐷 > 0,
namely
c) The equation (1) doesn’t have roots 𝐷 < 0 ⇔ 39𝑚2 −
60𝑚 − 28 >
Answer:
17. Let the set 𝐴 = {3,4,5} and 𝐵 = {4,5,6}. Write the elements
of the sets 𝐴2 ∩ 𝐵2 and (𝐴 ∖ (𝐵 ∖ 𝐴)) × (𝐵 ∩ 𝐴).
Solution. a) For the first set we have:
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Algebraic problems and exercises for high school
𝐴2 = {(3,3), (3,4), (3,5), (4,3), (4,4), (4,5), (5,3), (5,4), (5,5)},
𝐵2 = {(4,4), (4,5), (4,6), (5,4), (5,5), (5,6), (6,4), (6,5), (6,6)},
𝐴2 ∩ 𝐵2 = {(4,4), (4,5), (5,4), (5,5)}.
b) For the second set we have:
Then
Answer:
18. Given the sets 𝐴 = {1,2,3,4,5,6,7} and 𝐵 = {2,3,4}, solve
the equations 𝐴 ∆ 𝑋 = 𝐴 ∖ 𝐵 and (𝐴 ∆ 𝑌)∆ 𝐵 = 𝐶𝐴 (𝐵).
Solution. In order to solve the enunciated equations, we use the
properties of the symmetrical difference: 𝐴 ∆(𝐴 ∆ 𝐵) = 𝐵 , its
associativity and commutativity .
But
Consequently,
We calculate
Thus
Answer: 𝑋 = 𝐵, 𝑌 = ∅.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
19. The following sets are given
Determine the sets 𝐴 ∪ 𝐵, 𝐴 ∩ 𝐵, 𝐴, 𝐵, 𝐴 ∖ 𝐵, 𝐵 ∖ 𝐴, (𝐴 ∪ 𝐵) ∖
(𝐵 ∖ 𝐴) and Ā × (𝐵 ∖ 𝐴).
Solution. 1) We determine the sets 𝐴 and 𝐵.
The inequation (∗) is equivalent to the totality of the three
systems of inequations:
So
In other words,
2) We determine the required sets with the help of the graphical
representation
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Algebraic problems and exercises for high school
20. 40 students have a mathematics test paper to write, that
contains one problem, one inequation and one identity. Three
students have successfully solved only the problem, 5 only the
inequation, 4 have proven only the identity, 7 have solved not only the
problem, 6 students – not only the inequation, and 5 students have
proven not only the identity. The other students have solved
everything successfully. How many students of this type are there?
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
Solution. Let 𝐴 be the set of students who have correctly solved
only the problem, 𝐵 – only the inequation, 𝐶 – that have proven only
the identity, 𝐷 – the set of students that have solved only the problem
and the inequation, 𝐸 – the set of students that have solved only the
problem and have proven the identity, 𝐹 – the set of students who
have solved only the inequation and have proven the identity, and 𝐺 –
the set of students that have successfully solved everything.
From the given conditions, it follows that 𝑛 (𝐴) = 3, 𝑛 (𝐵) = 5,
𝑛 (𝐶) = 4, 𝑛 (𝐷) = 8, 𝑛 (𝐸) = 7, 𝑛 (𝐹) = 9.
But, as each of the students who have written the test paper
have solved at least one point of the test correctly and, because the
sets 𝐴, 𝐵, 𝐶, 𝐷, 𝐸, 𝐹, 𝐺 have only elements of the null set in common,
the union of the sets 𝐴, 𝐵, 𝐶, 𝐷, 𝐸, 𝐹, 𝐺 is the set of students that have
written the paper.
Consequently,
So
Answer: 4 students out of all that have taken the test have
solved everything successfully.
21. (Mathematician Dodjson’s problem)
In a tense battle, 72 out of 100 pirates have lost one eye, 75 –
one ear, 80 – one hand and 85 – one leg. What is the minimum number
of pirates that have lost their eye, ear, hand and leg at the same time?
Solution. We note with 𝐴 the set of one-eyed pirates, with 𝐵 –
the set of one eared pirates, with 𝐶 – the set of pirates with one hand
and with 𝐷 – the set of the pirates with one leg.
The problems requires to appreciate the set 𝐴 ∩ 𝐵 ∩ 𝐶 ∩ 𝐷.
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Algebraic problems and exercises for high school
It is obvious that the universal set 𝐸 is composed from the set
𝐴 ∩ 𝐵 ∩ 𝐶 ∩ 𝐷 and the number of pirates that have kept either both
eyes, or both ears, or both hands, or both legs.
Consequently,
It follows that the set 𝐸 is not smaller (doesn’t have fewer
̅ and 𝐴 ∩
elements) than the sum of the elements of the sets 𝐴̅, 𝐵̅, 𝐶̅ , 𝐷
𝐵 ∩ 𝐶 ∩ 𝐷 (equality would have been the case only if the sets 𝐴, 𝐵, 𝐶
and 𝐷, two by two, do not intersect).
But,
Substituting, we have 𝑛(𝐸) = 100 namely 100 ≤ 𝑛(𝐴 ∩ 𝐵 ∩
𝐶 ∩ 𝐷) + 30 + 25 + 20 + 15.
Consequently, 𝑛(𝐴 ∩ 𝐵 ∩ 𝐶 ∩ 𝐷) ≥ 100 − 30 − 25 − 20 = 10.
Therefore, no less than 10 pirates have lost their eye, ear, hand
and leg at the same time
Answer: No less than 10 pirates.
22. Of a total of 100 students, 28 study English, 8 – English and
German, 10 - English and French, 5 – French and German, 3 students all three languages. How many study only one language? How many
not even one language?
Solution. Let 𝐴 be the set of students that study English, 𝐵 German, 𝐶 - French.
Then the set of students attending both English and German is
𝐴 ∩ 𝐵, English and French - 𝐴 ∩ 𝐶, French and German - 𝐵 ∩ 𝐶,
English, German and French - 𝐴 ∩ 𝐵 ∩ 𝐶, and the set of students
that study at least one language is 𝐴 ∪ 𝐵 ∪ 𝐶.
From the conditions above, it follows that the students who only
study English form the set 𝐴\(𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶), only German 𝐵\(𝐴 ∩ 𝐵) ∪ (𝐵 ∩ 𝐶), only French - 𝐶\(𝐴 ∩ 𝐶) ∪ (𝐵 ∩ 𝐶).
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
But, because 𝐴 ∩ 𝐵 ⊆ 𝐴, we have 𝑛(𝐴\((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶))) =
𝑛(𝐴) − 𝑛((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)) = 𝑛(𝐴) − (𝑛(𝐴 ∩ 𝐵) + 𝑛(𝐴 ∩ 𝐶) −
𝑛(𝐴 ∩ 𝐵 ∩ 𝐶)) = 𝑛(𝐴) − 𝑛(𝐴 ∩ 𝐵) − 𝑛(𝐴 ∩ 𝐶) + 𝑛(𝐴 ∩ 𝐵 ∩ 𝐶).
Similarly,
𝑛 (𝐵\((𝐴 ∩ 𝐵) ∪ (𝐵 ∩ 𝐶))) = 𝑛(𝐵) − 𝑛(𝐴 ∩ 𝐵) −
𝑛(𝐵 ∩ 𝐶) + 𝑛(𝐴 ∩ 𝐵 ∩ 𝐶); 𝑛(𝐶\((𝐴 ∩ 𝐶) ∩ (𝐵 ∩ 𝐶)) = 𝑛(𝐶) −
𝑛(𝐴 ∩ 𝐶) − 𝑛(𝐵 ∩ 𝐶) + 𝑛(𝐴 ∩ 𝐵 ∩ 𝐶).
Let 𝐷 be the set of students who only study one language, then
Consequently,
= 28 + 30 + 42 − 2.8 − 2.10 − 2.5 + 3.3 = 63. 𝑛(𝐷) = 63.
The number of students who don’t study any language is equal
to the difference between the total number of students and the
number of students that study at least one language, namely 𝑛(𝐻) =
𝑛(𝑇) − 𝑛(𝐴 ∪ 𝐵 ∪ 𝐶), where 𝐻 is the set of students that don’t study
any language and 𝑇 the set of all 100 students.
From problem 20 we have
So 𝑛(𝐻) = 100 − 80 = 20.
Answer: 63 students study only one language, 20 students don’t
study any language.
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Algebraic problems and exercises for high school
1.3. Suggested exercises
1. Which of the following statements are true and which are
false?
2. Which of the following statements are true and which are
false (𝐴,𝐵 and 𝐶 are arbitrary sets)?
3. Let 𝐴 = {𝑥 ∈ ℚ│𝑥 2 − 12𝑥 + 35) = 0},
𝐵 = {𝑥 ∈ ℚ│𝑥 2 + 2𝑥 + 35) = 0};𝐶 = {𝑥 ∈ ℚ│𝑥 2 + 2𝑥 + 35) = 0}
a) Determine the sets 𝐴, 𝐵 and 𝐶.
b) State if the numbers 1/5, 5, 7, 1/2 belong to these sets or
not.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
4. Determine the sets:
5. Let 𝐴 = {𝑥 ∈ ℕ│x = 4n + 6m, n, m ∈ ℕ∗ };
a) Write three elements belonging to set 𝐴.
b) Determine if 26, 28, 33 belong to 𝐴.
6. Indicate the characteristic properties of the elements
belonging to the sets: 𝐴 = {4, 7, 10},
𝐵 = {3, 6, 12} , 𝐶 =
{1, 4, 9, 16, 25}, 𝐷 = {1, 8, 27, 64, 125}.
7. How many elements do the following sets have?
8. Let there be the set 𝐴 = {−3, −2, −1,1,2,3}. Determine the
subsets of 𝐴:
9. Determine the sets:
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Algebraic problems and exercises for high school
10. Compare (⊂, ⊃, =, ⊄, ⊅) the sets 𝐴 and 𝐵, if:
11. Let 𝐴 = {1, 2, 3, 4, 5, 6, 7}, 𝐵 = {5, 6, 7, 8, 9, 10}. Using the
symbols ∪,∩,∖, 𝐶 (complementary), express (with the help of 𝐴, 𝐵 and
ℕ∗ ) the sets:
.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
12. Determine the set 𝐸, in the case it is not indicated and its
parts 𝐴, 𝐵, 𝐶 simultaneously satisfy the conditions:
𝐴 has more elements than 𝐵;
𝐴 has more elements than 𝐵;
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Algebraic problems and exercises for high school
13. Determine the sets 𝐴, 𝐵, 𝐴 ∪ 𝐵, 𝐴 ∩ 𝐵, 𝐵 ∖ 𝐴, 𝐴 ∖ 𝐵, 𝐴∆𝐵,
𝐴 ∪ ( 𝐵 ∖ 𝐴), , 𝐴 ∖ (𝐵 ∖ 𝐴), 𝐴 × 𝐵, 𝐵 × 𝐴, (𝐴 ∪ 𝐵) × 𝐴, 𝐵 × ( 𝐴 ∖ 𝐵), if:
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
14. Let
a) Determine the sets:
b) How many elements does the set
𝐷 = {𝑥 ∈ 𝑀│𝑥 ≤
(699⁄100)} have?
15. Determine the sets 𝐴, 𝐵 ⊆ 𝐸, if
16. Determine the set 𝐸 and its parts 𝐴 and 𝐵, if
17. Compare the sets 𝐴 and 𝐵, if
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Algebraic problems and exercises for high school
18. Determine the values of the real parameter 𝑚 for which the
set 𝐴 has one element, two elements or it is void, if:
19. Determine the number of elements of the set 𝐴:
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
20. The sets 𝐴, 𝐵, 𝐶 are given. Determine 𝐴 ∩ 𝐵 ∩ 𝐶.
21. Determine 𝐴 ∩ 𝐵, if:
22. Prove the properties of the operations with sets.
23. Determine the equalities (𝐴, 𝐵, 𝐶 etc. are arbitrary sets):
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Algebraic problems and exercises for high school
24. Given the sets 𝐴 = {1, 2, 3}, 𝐵 = {2, 3, 4}, 𝐶 = {3, 4, 5} and
𝐷 = {4, 5, 6}, write the elements of the following sets:
25. Given the sets 𝐴, 𝐵 and 𝐶, solve the equations (𝐴𝑓𝐵)∆𝑋 =
𝐶, where 𝑓 ∈ {∪,∩,∖ ∆}:
̅ 𝐴 ∖ 𝐵, 𝐵 ∖ 𝐴, 𝐴 ∪
26. Determine the sets 𝐴 ∩ 𝐵, 𝐴 ∪ 𝐵, Ā, 𝐵,
(𝐵 ∖ Ā), 𝐴 ∩ (Ā ∖ 𝐵), if:
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
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Algebraic problems and exercises for high school
2. Relations, functions
2.1. Definitions and notations
Relations, types. Relations formation
Let 𝐴 and 𝐵 be two non-empty sets, and 𝐴 × 𝐵 their Cartesian
product. Any subset 𝑅 ⊆ 𝐴 × 𝐵 is called a relation between the
elements of 𝐴 and the elements of 𝐵. When 𝐴 = 𝐵, a relation like 𝑅 ⊆
𝐴 × 𝐴 is called a binary relation on set 𝐴.
If there exists a relation like 𝑅 ⊆ 𝐴 × 𝐵, then for an ordered pair
(𝑎, 𝑏) ∈ 𝐴 × 𝐵 we can have (𝑎, 𝑏) ∈ 𝑅 or (𝑎, 𝑏) ∉ 𝑅. In the first case,
we write 𝑎𝑅𝑏 and we read "𝑎 is in relation 𝑅 with 𝑏", and in the second
case – 𝑎𝑅𝑏, that reads "𝑎 is not in relation 𝑅 with 𝑏". We hold that
𝑎𝑅𝑏 ⟺ (𝑎, 𝑏) ∈ 𝑅.
By the definition domain of relation 𝑅 ⊆ A X B, we understand
the subset 𝔡𝑅 ⊆ 𝐴 consisting of only those elements of set 𝐴 that are in
relation 𝑅 with an element from 𝐵, namely
By values domain of relation 𝑅 ⊆ 𝐴 × 𝐵 we understand the
subset ρ𝑅 ⊆ 𝐵 consisting of only those elements of set 𝐵 that are in
relation 𝑅 with at least one element of 𝐴, namely
The inverse relation. If we have a relation 𝑅 ⊆ 𝐴 × 𝐵, then by
the inverse relation of the relation 𝑅 we understand the relation
𝑅 −1 ⊆ 𝐵 × 𝐴 as defined by the equivalence
namely
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
Example 1. Let us have 𝐴 = {1, 2}, 𝐵 = {4, 5, 6} and also the
relations
Determine the definition domain and the values domain of
these relations and their respective inverse relations.
Solution.
The formation of relations. Let 𝐴, 𝐵, 𝐶 be three sets and let us
consider the relations 𝑅 ⊆ 𝐴 × 𝐵, 𝑆 ⊆ 𝐵 × 𝐶. The relation 𝑅 ∘ 𝑆 ⊆ 𝐴 ×
𝐶 formed in accordance with the equivalence
namely
is called the formation of relation 𝑅 and 𝑆.
Example 2. Let us have 𝐴, 𝐵, 𝛼, 𝛽, 𝛾, from Example 1.
Determine
𝛾 ∘ 𝛽 −1 and (𝛽 ∘ 𝛾)−1.
Solution. We point out that the compound of the relations 𝛼 ⊆
𝐶 × 𝐷 with 𝛽 ⊆ 𝐸 × 𝐸 exists if and only if 𝐷 = 𝐸.
Because 𝛼 ⊆ 𝐴 × 𝐵, 𝛽 ⊆ 𝐴 × 𝐵, it follows that 𝛼 ∘ 𝛼 and 𝛼 ∘ 𝛽
doesn’t exist.
−1
We determine 𝛼 ∘ 𝛾:
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Algebraic problems and exercises for high school
(2,4) ∈ 𝛼 and {(4,1), (4,2)} ⊆ 𝛾 ⇒ {(2,1), (2,2)} ⊆ 𝛼 ∘ 𝛾;
(2,6) ∈ 𝛼 , but in 𝛾 we don’t have a pair with the first
component 6. It follows that
c) 𝛽 ⊆ 𝐴 × 𝐵 , 𝛾 ⊆ 𝐴 × 𝐵 ⇒ 𝛽 ∘ 𝛾 exists. By repeating the
reasoning from b),
and 𝛽 ∘ 𝛽 −1 = {(2,2)}.
and 𝛾 −1 ∘ 𝛽 −1 = {(1,2), (2,2)}.
The equality relation. Let 𝐴 be a set. The relation
is called an equality relation on 𝐴. Namely,
Of a real use is:
1stTheorem. Let 𝐴, 𝐵, 𝐶, 𝐷 be sets and 𝑅 ⊆ 𝐴 × 𝐵, 𝑆 ⊆ 𝐵 × 𝐶 ,
𝑇 ⊆ 𝐶 × 𝐷 relations. Then,
1) (𝑅 ∘ 𝑆) ∘ 𝑇 = 𝑅 ∘ (𝑆 ∘ 𝑇) (associativity of relations formation)
The equivalence Relations. The binary relation 𝑅 ⊆ 𝐴2 is called:
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
a) reflexive, if 𝑥𝑅𝑥 whatever 𝑥 ∈ 𝐴 is;
b) symmetrical, if (𝑥𝑅𝑦 ⇒ 𝑦𝑅𝑥), (⩝)𝑥, 𝑦 ∈ 𝐴;
c) transitive, if ((𝑥𝑅𝑦 ⋀ 𝑦𝑅𝑧) ⇒ 𝑥𝑅𝑧), (⩝) 𝑥, 𝑦, 𝑧 ∈ 𝐴 ;
d) anti-symmetrical, if ((𝑥𝑅𝑦 ⋀ 𝑦𝑅𝑥) ⇒ 𝑥 = 𝑦), (⩝) 𝑥, 𝑦, 𝑧 ∈ 𝐴;
e) equivalence relation on A, if it is reflexive, symmetrical and
transitive;
f) irreflexive, if 𝑥𝑅𝑥 whatever 𝑥 ∈ 𝐴 is.
Let 𝑅 be an equivalence relation on set 𝐴. For each element 𝑥 ∈
𝐴, the set
is called the equivalence class of 𝑥 modulo 𝑅 (or in relation to R), and
the set
is called a factor set (or quotient set) of 𝐴 through 𝑅.
The properties of the equivalence classes. Let 𝑅 be an
equivalence relation on set 𝐴 and 𝑥, 𝑦 ∈ 𝐴 . Then, the following
affirmations have effect:
Partitions on a set. Let 𝐴 be a non-empty set. A family of
subsets {𝐴𝑖 │𝑖 ∈ 𝐼} of 𝐴 is called a partition on 𝐴 (or of 𝐴), if the
following conditions are met:
2nd Theorem. For any equivalence relation 𝑅 on set 𝐴, the factor
set 𝐴/ 𝑅 = {𝑅𝑥 │𝑥 ∈ 𝐴} is a partition of 𝐴.
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Algebraic problems and exercises for high school
3rd Theorem. For any partition 𝑆 = {𝐴𝑖 │𝑖 ∈ 𝐼} of 𝐴, there exists
a sole equivalence relation 𝛼𝑆 on 𝐴, hence
The relation 𝛼𝑆 ⊆ 𝐴 is formed according to the rule
It is easily established that 𝛼𝑆 is an equivalence relation on 𝐴 and
the required equality.
Example 3. We define on set ℤ the binary relation 𝛼 according to
the equivalence 𝑎𝛼𝑏 ⇔ (𝑎 − 𝑏) ⋮ 𝑛, where 𝑛 ∈ ℕ∗ , 𝑛 fixed, (⩝)𝑎, 𝑏 ∈
ℤ.
a) Prove that 𝛼 is an equivalence relation on ℤ.
b) Determine the structure of the classes of equivalence.
c) Form the factor set ℤ/𝛼. Application: 𝑛 = 5.
Solution. a) Let 𝑎, 𝑏, 𝑐 ∈ ℤ. Then:
1) reflexivity 𝑎 − 𝑎 = 𝑂 ⋮⇒ 𝑛 𝑎𝛼𝑎;
2) symmetry
3) transitivity
From 1) - 3) it follows that 𝛼 is an equivalence relation on ℤ.
b) Let 𝑎 ∈ ℤ. Then
In accordance with the theorem of division with remainder for
integer numbers 𝑎 and 𝑛, we obtain
Then
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
where 𝑠 = 𝑡 + 𝑞 ∈ ℤ, and therefore
where 𝑟𝑎 is the remainder of the division of 𝑎 through 𝑛. But,
In other words, the equivalence class of 𝑎 ∈ ℤ coincides with the
class of the remainder of the division of 𝑎 through 𝑛.
c) Because by the division to 𝑛 only the remainders 0, 1, 2, . . .,
𝑛 − 1, can be obtained, from point b) it follows that we have the
exact 𝑛 different classes of equivalence:
The following notation is customarily used
. Then
where 𝑖̂ consists of only those integer numbers that divided by 𝑛 yield
the remainder
.
For n = 5, we obtain
with
Definition. The relation 𝛼 is called a congruency relation
modulo 𝑛 on ℤ, and class â = 𝛼𝑎 is called a remainder class modulo
𝑛 and its elements are called the representatives of the class.
The usual notation:
𝑎𝛼𝑏 ⇔ (𝑎 − 𝑏) ⋮ 𝑛 ⇔ 𝑎 ≡ 𝑏 (mode 𝑛) (𝑎 is congruent with 𝑏
modulo 𝑛),and
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Algebraic problems and exercises for high school
is the set of all remainder classes modulo 𝑛.
Example 4 (geometric). Let 𝜋 be a plane and 𝐿 the set of all the
straight lines in the plane. We define 𝐿 as the binary relation 𝛽 in
accordance with
a) Show that 𝛽 is an equivalence relation on 𝐿.
b) Describe the equivalence classes modulo 𝛽.
c) Indicate the factor set.
Solution. a) It is obvious that 𝛽 is an equivalence relation (each
straight line is parallel to itself; if 𝑑1 ∥ 𝑑2 ⇒ 𝑑2 ∥ 𝑑1 and
b) Let 𝑑 ∈ 𝐿. Then the class
consists only of those straight lines from 𝐿 that are parallel with the
straight line 𝑑.
c) 𝐿/𝛽 = {𝛽𝑑 │𝑑 ∈ 𝐿} is an infinite set, because there are an
infinity of directions in plane 𝜋.
Example 5. The set𝐴 = {1,2,3,4,5,6, 7,8,9}is given and its parts
a) Show that 𝑆 = {𝐴1 , 𝐴2 , 𝐴3 , 𝐴4 } is a partition of 𝐴.
b) Determine the equivalence relation 𝛼𝑆 on 𝐴.
c) Is 𝑇 = {𝐵1 , 𝐵2 , 𝐵3 } a partition on 𝐴?
Solution. a) 1) We notice that
meaning the system 𝑆 of the subsets of the set A defines a partition on
𝐴.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
b) In accordance to the 3rd Theorem, we have
So:
𝛼𝑆
= {( 1,1), (1,2), (2,1), (2,2), (3,3), (3,4), (3,6), (4,3), (4,4), (4,6), (6,3),
(6,4), (6,6), (5,5), (5,7), (5,8), (8,5), (8,8), (8,7), (7,5), (7,7), (7,8), (9,9)}.
which proves that system 𝑇 does not define a partition on 𝐴.
Order relations. A binary relation 𝑅 on the set 𝐴 is called an
order relation on 𝐴, if it is reflexive, anti-symmetrical and transitive.
If 𝑅 is an order relation on 𝐴, then 𝑅 −1 is also an order relation
on 𝐴 (verify!). Usually, the relation 𝑅 is denoted by "≤" and the
relation 𝑅 −1 by "≥", hence
With this notation, the conditions that "≤" is an order relation
on the set 𝐴 are written:
 reflexivity 𝑥 ∈ 𝐴 ⇒ 𝑥 ≤ 𝑥;
 asymmetry (𝑥 ≤ 𝑦⋀𝑦 ≤ 𝑥) ⇒ 𝑥 = 𝑦;
 transitivity (𝑥 ≤ 𝑦⋀𝑦 ≤ 𝑧) ⇒ 𝑥 ≤ 𝑧.
Example 6. On the set ℕ we define the binary relation 𝛾 in
accordance with
Show that 𝛾 is an order relation on ℕ.
Solution. We verify the condition from the definition of order
relations.
1) Reflexivity
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Algebraic problems and exercises for high school
2) Asymmetry. Let 𝑎, 𝑏 ∈ ℕ with 𝑎𝛾𝑏 and 𝑏𝛾𝑎. It follows that
there are the natural numbers 𝑐, 𝑑 ∈ ℕ with 𝑎 = 𝑏 . 𝑐 and 𝑏 = 𝑎 . 𝑑.
Then
which implies that
3) Transitivity. Let 𝑎, 𝑏, 𝑐 ∈ ℕ with 𝑎𝛾𝑏 and 𝑏𝛾𝑐. Then there
exists 𝑢, 𝑣 ∈ ℕ with 𝑎 = 𝑏𝑢 and 𝑏 = 𝑐𝑣 hence,
Because 𝑢, 𝑣 ∈ ℕ, the implication is true
which proves that 𝛾 is an order relation on the set ℕ .
Remark. The order relation 𝛾 is called a divisibility relation on ℕ
and it is written as 𝑎 ⋮ 𝑏, namely
(𝑏│𝑎 reads "b divides a", and 𝑎 ⋮ 𝑏 "a is divisible through b").
Functional relations
Let 𝐴 and 𝐵 be two sets. A relation 𝑅 ⊆ 𝐴 × 𝐵 is called an
application or a functional relation, if the following conditions are met:
1)⩝ 𝑥 ∈ 𝐴(∃)𝑦 ∈ 𝐵, so as 𝑥𝑅𝑦;
An application (or function) is a triplet 𝑓 = (𝐴, 𝐵, 𝑅), where 𝐴
and 𝐵 are two sets and 𝑅 ⊆ 𝐴 × 𝐵 is a functional relation.
If 𝑅 ⊆ 𝐴 × 𝐵 is an application, then for each 𝑥 ∈ 𝐴 there exists,
according to the conditions 1) and 2) stated above, a single element
𝑦 ∈ 𝐵, so that 𝑥𝑅𝑦; we write this element 𝑦 with 𝑓(𝑥). Thus,
The element 𝑓(𝑥) ∈ 𝐵 is called the image of element 𝑥 ∈ 𝐴
through application 𝑓, set 𝐴 is called the definition domain of 𝑓 noted
by 𝐷(𝑓) = 𝐴, and set 𝐵 is called the codomain of 𝑓 and we usually say
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
that 𝑓 is an application defined on 𝐴 with values in 𝐵. The functional
relation 𝑅 is also called the graphic of the application (function) 𝑓,
denoted, later on, by 𝐺𝑓 . To show that 𝑓 is an application defined on 𝐴
𝑓
with values in 𝐵 , we write 𝑓: 𝐴 ⟶ 𝐵 or 𝐴 → 𝐵 , and, instead of
describing what the graphic of 𝑅 (or 𝑓′𝑠) is, we indicate for each 𝑥 ∈ 𝐴
its image 𝑓(𝑥). Then,
i.e.
The equality of applications. Two applications 𝑓 = (𝐴, 𝐵, 𝑅)
and 𝑔 = (𝐶, 𝐷, 𝑆) are called equal if and only if they have the same
domain 𝐴 = 𝐶, the same codomain 𝐵 = 𝐷 and the same graphic
𝑅 = 𝑆. If 𝑓, 𝑔: 𝐴 ⟶ 𝐵, the equality 𝑓 = 𝑔 is equivalent to 𝑓(𝑥) =
𝑔(𝑥), (⩝)𝑥 ∈ 𝐴, that is to say
The identical application. Let 𝐴 be a set. The triplet (𝐴, 𝐴, 1𝐴 ) is,
obviously, an application, denoted by the same symbol 1𝐴 (or 𝜀) and it
is called the identical application of set 𝐴.
We have
Consequently,
1𝐴 : 𝐴 ⟶ 𝐴 and 1𝐴 (𝑥) = 𝑥 for (⩝)𝑥 ∈ 𝐴.
By 𝐹(𝐴, 𝐵) we will note the set of functions defined on 𝐴 with
values in 𝐵. For 𝐵 = 𝐴, we will use the notation 𝐹(𝐴) instead of
𝐹(𝐴, 𝐴).
In the case of a finite set 𝐴 = {𝑎1 , 𝑎2 , … , 𝑎𝑛 }, a function
𝜑: 𝐴 ⟶ 𝐴 is sometimes given by means of a table as is:
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Algebraic problems and exercises for high school
where we write in the first line the elements 𝑎1 , 𝑎2 , … , 𝑎𝑛 of set 𝐴, and
in the second line their corresponding images going through 𝜑, namely
𝜑(𝑎1 ), 𝜑(𝑎2 ), … , (𝑎𝑛 ).
When 𝐴 = {1,2, … , 𝑛}, in order to determine the application,
we will use 𝜑: 𝐴 ⟶ 𝐴 and the notation
more frequently used to write the bijective applications of the set 𝐴 in
itself. For example, if 𝐴 = {1,2}, then the elements of 𝐹(𝐴) are:
If 𝑓: 𝐴 ⟶ 𝐵, 𝑋 ⊆ 𝐴, 𝑌 ⊆ 𝐵, then we introduce the notations:
- the image of subset 𝑋 through the application 𝑓.
Particularly, 𝜑(𝐴) = 𝐼𝑚𝜑 , the image of application 𝜑 ;
is the preimage of subset 𝑌 through the application 𝑓.
Particularly, for 𝑦 ∈ 𝐵, instead of writing 𝑓 −1 ({𝑦}), we simply
note 𝑓 −1 (𝑦), that is
- the set of all preimages of 𝑦 through the application 𝑓, and
- the complete preimage of set 𝐵 through the application 𝑓.
The formation of applications. We consider the applications
𝑓 = (𝐴, 𝐵, 𝑅) and 𝑔 = (𝐵, 𝐶, 𝑆), so as the codomain of 𝑓 to coincide
with the domain of 𝑔. We form the triplet 𝑔 ∘ 𝑓 = (𝐴, 𝐶, 𝑅 ∘ 𝑆).
Then 𝑔 ∘ 𝑓 is also an application, named the compound of the
application 𝑔 with the application 𝑓, and the operation "∘" is called the
formation of applications. We have
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
namely
4th Theorem. Given the applications
. It follows that a) (ℎ ∘ 𝑔) ∘ 𝑓 = ℎ ∘ (𝑔 ∘ 𝑓)
(associativity of the formation of applications); b) 𝑓 ∘ 1𝐴 = 1𝐵 ∘ 𝑓 = 𝑓.
Example 7. We consider the relations 𝑅 ⊆ ℝ × ℝ and 𝑆 ⊆
[0, +∞) × [0, +∞), 𝑇 ⊆ ℝ × ℝ, defined as it follows: 𝑥𝑅𝑦 ⇔ 𝑥 2 = 𝑦,
𝑥𝑆𝑦 ⇔ 𝑥 = 𝑦 2, 𝑥𝑇𝑦 ⇔ 𝑦 = 𝑥 + 1.
A. Determine which of the relations 𝑅, 𝑅 −1, 𝑆, 𝑆 −1, 𝑇, 𝑇 −1 are
functional relations.
B. Find the functions determined at point A.
C. Calculate 𝑓 ∘ 𝑔, 𝑔 ∘ 𝑓, 𝑓 ∘ ℎ, ℎ ∘ 𝑓, ℎ ∘ ℎ−1 , ℎ−1 ∘ ℎ (f, g, h are
the functions from point B.)
D. Calculate (𝑓 ∘ ℎ)( −3), (ℎ−1 ∘ ℎ)(1/2), (ℎ ∘ 𝑓)(1/3).
Solution. We simultaneously solve 𝐴 and 𝐵.
a) We examine the relation 𝑅.
𝑥 ∈ ℝ ⇒ 𝑥 2 ∈ ℝ ⇒ 𝑦 ∈ ℝ, i.e.
1)⩝ 𝑥 ∈ ℝ(∃)𝑦 = 𝑥 2 ∈ ℝ, so that 𝑥𝑅𝑦;
which means that 𝑅 is a functional relation. We write the application
determined by 𝑅 through 𝑓, 𝑓 = ( ℝ, ℝ, 𝑅).
b) We examine the relation 𝑅 −1 .
which means that if 𝑥 ∈ ℝ = {𝑎 ∈ ℝ│𝑎 < 0}, then there is no 𝑥 ∈ ℝ,
so that 𝑦𝑅 −1 𝑥 , which proves that 𝑅 −1 is not a functional relation.
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Algebraic problems and exercises for high school
c) For the relation 𝑆, we have: 𝑆 and 𝑆 −1 are functional relations.
We denote by 𝑔 = ([0, +∞), [0, +∞), 𝑆),and 𝑔−1 = ([0, +∞), [0, +∞),
𝑆 −1 ), the functions (applications) defined by 𝑆 and 𝑆 −1, respectively.
d) We examine the relation 𝑇:
1) 𝑥 ∈ ℝ ⇒ 𝑥 + 1 ∈ ℝ and so
so that 𝑥𝑇𝑦;
which proves that 𝑇 is a functional relation.
e) For the relation 𝑇 −1, we obtain:
1)(⩝)𝑥 ∈ ℝ (∃)𝑥 = 𝑦 − 1 ∈ ℝ, so that y𝑇 −1 𝑥;
meaning that 𝑇 −1 is also a functional relation. We denote by ℎ =
(ℝ, ℝ, 𝑇) and ℎ−1 = (ℝ, ℝ, 𝑇 −1 ).
C. From A and B, it follows that:
a) Then 𝑓 ∘ 𝑔, 𝑔 ∘ 𝑓 and 𝑔 ∘ ℎ don’t exist, because the domains
and codomains do not coincide:
𝑓 = (ℝ, ℝ, 𝑅), 𝑔 = ([0, +∞), [0, +∞), 𝑆), ℎ = (ℝ, ℝ, 𝑇)).
b) We calculate 𝑓 ∘ ℎ, ℎ ∘ 𝑓, ℎ ∘ ℎ−1 and ℎ−1 ∘ ℎ.
So 𝑓 ∘ ℎ ≠ ℎ ∘ 𝑓 and ℎ ∘ ℎ−1 = ℎ−1 ∘ ℎ = 1ℝ .
D. We calculate:
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
Injective, surjective and bijective applications. An application
𝑓: 𝐴 ⟶ 𝐵 is called:
1) injective, if
(equivalent: 𝑎1 ≠ 𝑎2 ⇒ 𝑓(𝑎1) ≠ 𝑓(𝑎2 ));
2) surjective, if
(any element from B has at least one preimage in 𝐴);
3) bijective, if 𝑓 is injective and surjective.
Remark. To prove that a function 𝑓: 𝐴 ⟶ 𝐵 is a surjection, the
equation 𝑓( 𝑥) = 𝑏 must be solvable in 𝐴 for any 𝑏 ∈ 𝐵.
Of a real use are:
𝑓
𝑔
5th Theorem. Given the applications 𝐴 → 𝐵 → 𝐶, we have:
a) if 𝑓 and 𝑔 are injective, then 𝑔 ∘ 𝑓 is injective;
b) if 𝑓 and 𝑔 are surjective, then 𝑔 ∘ 𝑓 is surjective;
c) if 𝑓 and 𝑔 are bijective, then 𝑔 ∘ 𝑓 is bijective;
d) if 𝑔 ∘ 𝑓 is injective, then 𝑓 is injective;
e) if 𝑔 ∘ 𝑓 is surjective, then 𝑔 is surjective.
6th Theorem. The application 𝑓: 𝐴 ⟶ 𝐵 with the graphic 𝐺𝑓 =
𝑅 is a bijective application, if and only if the inverted relation 𝑅 −1 is a
functional relation (f −1is an application).
This theorem follows immediately from
Inverted application. Let 𝑓: 𝐴 ⟶ 𝐵 be a bijective application
with the graphic 𝐺𝑓 = 𝑅. From the 6th Theorem, it follows that the
triplet 𝑓 −1 = (𝐵, 𝐴, 𝑅 −1 ) is an application (function). This function is
called the inverse of function 𝑓.
We have
𝑓 −1 : 𝐵 ⟶ 𝐴, and for 𝑦 ∈ 𝑅,
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Algebraic problems and exercises for high school
i.e.
7th Theorem. Application 𝑓: 𝐴 ⟶ 𝐵 is bijective, if and only if
there exists an application 𝑔: 𝐵 ⟶ 𝐴 with 𝑔 ∘ 𝑓 = 1𝐴 and 𝑓 ∘ 𝑔 = 1𝐵 .
In this case, we have 𝑔 = 𝑓 −1.
Real functions
The function 𝑓: 𝐴 ⟶ 𝐵 is called a function of real variable, if
𝐴 = 𝐷(𝑓) ⊆ ℝ. The function of real variable 𝑓: 𝐴 ⟶ 𝐵 is called a real
function, if 𝐵 ⊆ ℝ. In other words, the function 𝑓: 𝐴 ⟶ 𝐵 is called a
real function, if A ⊆ ℝ and B ⊆ ℝ . The graphic of the real
function 𝑓: 𝐴 ⟶ 𝐵 is the subset 𝐺𝑓 of ℝ2 , formed by all the
pairs (𝑥, 𝑦) ∈ ℝ2 with 𝑥 ∈ 𝐴 and 𝑦 = 𝑓(𝑥), i.e.:
If the function 𝑓 is invertible, then
Traditionally, instead of 𝑓 −1 (𝑦) = 𝑥 we write 𝑦 = 𝑓 −1 (𝑥). Then
the graphic of the inverted function 𝑓 −1 is symmetrical to the graphic
function 𝑓 in connection to the bisector 𝑦 = 𝑥.
Example 8. For the function
the inverted function is
The graphic of function 𝑓 is the branch of the parabola 𝑦 = 𝑥 2
comprised in the quadrant I, and the graphic of function 𝑓 −1 is the
branch of parabola 𝑥 = 𝑦 2 comprised in the quadrant I (see below Fig.
2.1).
Algebra operations with real functions
Let 𝑓, 𝑔: 𝐷 ⟶ ℝ be two real functions defined on the same set.
We consider the functions:
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
𝑠 = 𝑓 + 𝑔: 𝐷 ⟶ ℝ ,
defined through 𝑠(𝑥) = 𝑓(𝑥) + 𝑔(𝑥), (⩝)𝑥 ∈ 𝐷;
𝑠 = 𝑓 + g – sum-function.
𝑝 = 𝑓 ∙ 𝑔: 𝐷 ⟶ ℝ ,
defined through 𝑝(𝑥) = 𝑓(𝑥) · 𝑔(𝑥), (⩝)𝑥 ∈ 𝐷;
𝑝 = 𝑓 . 𝑔 – product-function.
𝑑 = 𝑓 − 𝑔: 𝐷 ⟶ ℝ,
defined through (𝑥) = 𝑓(𝑥) − 𝑔(𝑥), (⩝)𝑥 ∈ 𝐷;
𝑑 = 𝑓 − 𝑔 – difference-function.
𝑓
𝑞 = 𝑔 : 𝐷1 ⟶ ℝ,
𝑓(𝑥)
defined through 𝑞(𝑥) = 𝑔(𝑥) , (⩝)𝑥 ∈ 𝐷1
where 𝐷1 = {𝑥 ∈ 𝐷│𝑔(𝑥) ≠ 𝑂};
𝑓
𝑞 = 𝑔 – quotient-function.
│𝑓│: 𝐷 ⟶ ℝ,
𝑓(𝑥), 𝑖𝑓 𝑓(𝑥) ≥ 0,
}
defined through │𝑓│ (𝑥) = │𝑓 (𝑥) = {
−𝑓(𝑥), 𝑖𝑓 𝑓(𝑥) < 0,
for (⩝)𝑥 ∈ 𝐷;
│𝑓│- module-function.
Figure 2.1.
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Algebraic problems and exercises for high school
Example 9. Let 𝑓, 𝑔: [0, +∞) ⟶ ℝ with 𝑓(𝑥) = 𝑥 2 and 𝑔(𝑥) =
√𝑥. Determine 𝑓 ± 𝑔, 𝑓 ∙ 𝑔 and 𝑓/𝑔·
Solution. As 𝑓, 𝑔 share the same definition domain, the
functions 𝑓 ± 𝑔, 𝑓 ∙ 𝑔 and 𝑓/𝑔 make sense and
Example 10. Let 𝑓: ℝ ⟶ [0, +∞), 𝑓(𝑥) = 𝑥 2and 𝑔: [0, +∞) ⟶
ℝ, 𝑔(𝑥) = √𝑥. Determine a) 𝑔 ∘ 𝑓 and b) 𝑓 ∘ 𝑔.
Solution. a) 𝑔 ∘ 𝑓 = 𝜑 makes sense and we obtain the function
𝜑: ℝ ⟶ ℝ with
b) 𝜓 = 𝑓 ∘ 𝑔: [0, +∞) ⟶ [0, +∞), also
makes
sense,
with
2.2. Solved exercises
1. Let 𝐴 = {2,4,6,8} and 𝐵 = {1,3,5,7},
{(𝑥, 𝑦)|𝑥 ≥ 6 𝑜𝑟 𝑦 ≤ 1} :
a) What is the graphic of the relation 𝛼?
b) Make the schema of the relation 𝛼.
Solution.
a) Because 𝑥 ∈ 𝐴, and 𝑦 ∈ 𝐵, it follows that
So:
b) See Fig. 2.2 below.
59
𝑎 ⊆ 𝐴 × 𝐵, 𝛼 =
Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
Figure 2.2.
2. Let 𝐴 = {1,2,3,4} and 𝐵 = {1,3,5,7,8}. Write the graphic of
the relation 𝛼 = {(𝑥, 𝑦)│3𝑥 + 4𝑦 = 10} ⊆ 𝐴 × 𝐵.
Solution. We observe that 3𝑥 + 4𝑦 = 10 ⇔ 3𝑥 = 2(5 −
2𝑦) ⇒ 𝑥 is even and (5 − 2𝑦) ⋮ 3. Considering that 𝑥 ∈ 𝐴, and 𝑦 ∈ 𝐵,
we obtain: 𝑥 ∈ (2,4), and 𝑦 ∈ (1,7). The equality 3𝑥 + 4𝑦 = 10 is
met only by 𝑥 = 2 and 𝑦 = 1. So 𝐺𝛼 = {(2,1)}.
3. Let 𝐴 = {1,3,4,5} and 𝐵 = {1,2,5,6}. Write the relation 𝛼
using letters 𝑥 ∈ 𝐴, and 𝑦 ∈ 𝐵, if the graphic of the relation 𝛼 is known.
Solution. (𝑥, 𝑦) ∈ 𝐺𝛼 ⇔ 𝑦 ⋮ 𝑥.
4. On the natural numbers set ℕ we consider the
relations 𝛼, 𝛽, 𝛾, 𝜔 ⊆ ℕ2 , defined as it follows: (𝑥, 𝑦 ∈ ℕ): 𝛼 =
{(3,5), (5,3), (3,3), (5,5)} ; 𝑥𝛽𝑦 ⇔ 𝑥 ≤ 𝑦 ; 𝑥𝛾𝑦 ⇔ 𝑦 − 𝑥 = 12 ;
𝑥𝜔𝑦 ⇔ 𝑥 = 3𝑦.
a) Determine 𝛿𝛼 , 𝜌𝛼 , 𝛿𝛽 , 𝜌𝛽 , 𝛿𝛾 , 𝜌𝛾 , 𝛿𝜔 , 𝜌𝜔 .
b) What properties do each of the relations 𝛼, 𝛽, 𝛾, 𝜔 have?
c) Determine the relations 𝛼 −1 , 𝛽 −1 , 𝛾 −1 and 𝜔−1 .
d) Determine the relations 𝛽 ∘ 𝛾, 𝛾 ∘ 𝛽, 𝛾 −1 ∘ 𝛽 −1 , (𝛽 ∘ 𝛾)−1 ,
𝛾 ∘ 𝜔 and 𝜔−1 ∘ 𝜔.
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Algebraic problems and exercises for high school
Solution:
(for any natural numbers, there are natural numbers that exceed it).
(for any natural numbers there is at least one natural number that
exceeds it).
(for example, equation 2 = 𝑥 + 12 doesn’t have solutions in ℕ).
(for any 𝑛 ∈ ℕ, we have 𝑛 = 3𝑛/𝑛).
b) 1) 𝛼 is symmetrical, transitive, but is not reflexive. For
example (2,2) ∈ 𝛼 ; because (3,3) ∈ 𝛼 :, it follows that 𝛼 isn’t
antireflexive either.
2) 𝛽 is
- reflexive(𝑥 ≤ 𝑥, (⩝) 𝑥 ∈ ℕ),
- anti-symmetrical ((𝑥𝛽𝑦 ⋀ 𝑦𝛽𝑥) ⇒
(𝑥 ≤ 𝑦 ⋀ 𝑦 ≤ 𝑥) ⇒ 𝑥 = 𝑦),
- transitive ((𝑥 ≤ 𝑦 ⋀ 𝑦 ≤ 𝑧) ⇒ 𝑥 ≤ 𝑧).
So 𝛽 is an order relation on set ℕ.
3) 𝛾 is antireflexive (𝑥 ≤ 𝑥, (⩝) 𝑥 ∈ ℕ), anti-symmetrical
((𝑥𝛾𝑦 ∧ 𝑦𝛾𝑥) ⇒ (𝑦 − 𝑥 = 12 and 𝑥 − 𝑦 = 12) ⇒ 𝑥 − 𝑦 = 𝑦 − 𝑥 ⇒
𝑥 = 𝑦), but it is not symmetrical (𝑥𝛾𝑦 ⇒ 𝑦 − 𝑥 = 12 ⇒ 𝑥 − 𝑦 =
−12 ⇒ 𝑥𝛾𝑦), and it is not transitive ((𝑥𝛾𝑦 ⋀ 𝑦𝛾𝑧) ⇒ (𝑦 − 𝑥 =
12 ⋀ 𝑧 − 𝑦 = 12) ⇒ 𝑧 − 𝑥 = 24 ⇒ 𝑥𝛾𝑧 ).
4) 𝜔 is not antireflexive (𝑥 ≠ 3𝑥, (⩝) 𝑥 ∈ ℕ∗ , 0 = 3.0), it is antisymmetrical ((𝑥𝜔𝑦 ⋀ 𝑦𝜔𝑥) ⇒ (𝑥 = 3𝑦 ⋀ 𝑦 = 3𝑥) ⇒ 𝑥 = 9𝑥 ⇒ 𝑥 =
0 = 𝑦 ), and for 𝑥 ≠ 0 ≠ 𝑦, the equalities 𝑥 = 3𝑦 and 𝑦 = 3𝑥 cannot
take place), it is not reflexive (for example, 1 ≠ 3.1 ⇒ 1𝜔1), it is not
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
transitive ((𝑥𝜔𝑦 ⋀ 𝑦𝜔𝑧) ⇒ (𝑥 = 3𝑦 ⋀ 𝑦 = 3𝑧) ⇒ 𝑥 = 9𝑧 ≠ 3𝑧 ,
generally ⇒ 𝑥𝜔𝑧).
So
i.e.
i.e.
i.e.
which implies
i.e.
In other words,
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Algebraic problems and exercises for high school
⇔ 𝑥 + 12 = 3𝑦, which implies
which implies
5. We consider the binary relation defined on ℝ as it follows:
a) Prove that 𝛼 is an equivalence relation.
b) Determine the factor set ℝ/𝛼.
Solution.
1) Because 𝑥 = 𝑥, (⩝) 𝑥 ∈ ℝ, we have 𝑥𝛼𝑥, (⩝) 𝑥 ∈ ℝ.
3) Let 𝑎𝛼𝑏 and 𝑏𝛼𝑐, 𝑎, 𝑏, 𝑐 ∈ ℝ. Then
In other words, the binary relation 𝛼 is reflexive, symmetrical
and transitive, which proves that 𝛼 is an equivalence relation on ℝ.
b) Let 𝛼 ∈ ℝ. We determine the class 𝛼𝑎 of equivalence of 𝑎 in
connection to 𝛼.
Then the set factor is
We observe that
In other words, for 𝑎 ≠ 1, we have 𝑎 ≠ 2 − 𝑎 and therefore
𝛼𝑎 = 𝛼2−𝑎 , and the factor can also be written
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
6. Let 𝐴 = {1,2,3,4,5} and 𝐵 = {𝑎, 𝑏, 𝑐, 𝑑, 𝑒}. Which of the
diagrams in Fig. 2.3 represents a function defined on 𝐴 with values in 𝐵
and which doesn’t?
Solution. The diagrams a), c), e) reveal the functions defined on
𝐴 with values in 𝐵. Diagram b) does not represent a function defined
on 𝐴 with values in 𝐵, because the image of 2 is not indicated.
Diagram d) also does not represent a function defined on 𝐴 with values
in 𝐵, because 1 has two corresponding elements in 𝐵, namely 𝑎, 𝑑.
7. Let A = {1, 2, 3} and 𝐵 = {𝑎, 𝑏}. Write all the functions
defined on 𝐴 with values in 𝐵, indicating the respective diagram.
Solution. There are eight functions defined on 𝐴 with values in 𝐵.
Their diagrams are represented in Fig. 2.4.
Figure 2.3.
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Algebraic problems and exercises for high school
Figure 2.4.
8. Let 𝐴 = {1, 2, 3, 4}, 𝐵 = {𝑎, 𝑏, 𝑐}, 𝐶 = {𝑥, 𝑦, 𝑧, 𝑡} and 𝐷 =
{1, 2}.
a) Draw the corresponding diagrams for two surjective functions
defined on 𝐴 with values in𝐵.
b) Draw the diagram of the injective functions defined on 𝐷 with
values in 𝐵.
c) Draw the diagram of a function denied on 𝐵 with values in 𝐶,
that is not injective.
d) Draw the corresponding diagrams for two bijective functions
defined on 𝐴 with values in 𝐶.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
Solution. a) The diagrams of two surjective and two nonsurjective functions defined on 𝐴 with values in 𝐵 are represented in
Fig. 2.5 a) and b).
Figure 2.5.
b) The diagram of the injective functions defined on 𝐷 with
values in 𝐵 are represented in Fig. 2.6.
c) The diagram of a non-injective function defined on 𝐵 with
values in 𝐶 is represented in Fig. 2.7.
d) The diagrams of the two bijective functions defined on 𝐴 with
values in 𝐶 are represented in Fig. 2.8.
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Algebraic problems and exercises for high school
Figure 2.5.
Figure 2.6.
Figure 2.7.
Figure 2.8.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
9. Let 𝐴 = {0, 1,5, 6}, 𝐵 = {0, 3, 8, 15} and the function
𝑓: 𝐴 ⟶ 𝐵 as defined by the equality 𝑓( 𝑥) = 𝑥 2 − 4𝑥 + 3, (⩝) 𝑥 ∈
𝐴.
a) Is the function 𝑓 surjective?
b) Is 𝑓 injective?
c) Is 𝑓 bijective?
Solution. a) We calculate 𝑓(𝐴) = {𝑓(0), 𝑓(1), 𝑓(5), 𝑓(6)} =
{3,0, 8, 15} = 𝐵. So, 𝑓 is surjective.
b) The function 𝑓 is also injective, because 𝑓(0) = 3, 𝑓(1) = 0,
𝑓(5) = 8 and 𝑓(6) = 15, i.e. 𝑥1 ≠ 𝑥2 ⇒ 𝑓(𝑥1 ) ≠ 𝑓(𝑥2 )
c) The function 𝑓 is bijective.
10. Using the graphic of the function 𝑓: 𝐴 ⟶ 𝐵, 𝑓: ℝ ⟶ ℝ,
show that the function 𝑓 is injective, surjective or bijective.
𝑥 − 2, 𝑖𝑓 𝑥 ∈ (−∞, 2),
a) 𝑓(𝑥) = {
3𝑥 − 6, 𝑖𝑓 𝑥 ∈ [2, +∞).
−3𝑥, 𝑖𝑓 − 1 < 𝑥 ≤ 0,
[−1,
c) 𝑓:
+∞] ⟶ ℝ, 𝑓(𝑥) = { −𝑥, 𝑖𝑓 0 < 𝑥 ≤ 1,
−0.5(𝑥 + 1), 𝑖𝑓 𝑥 > 1.
Solution. If any parallel to the axis of the abscises cuts the
graphic of the function in one point, at most (namely, it cuts it in one
point or not at all), then the function is injective. If there is a parallel to
the abscises’ axis that intersects the graphic of the function in two or
more points, the function in not injective. If 𝐸(𝑓) is the values set of
the function 𝑓 and any parallel to the abscises’ axis, drawn through the
ordinates axis that are contained in 𝐸(𝑓), cuts the graphic of the
function 𝑓 in at least one point, the function is surjective. It follows that
a function 𝑓 is bijective if any parallel to the abscises’ axis, drawn
through the points of 𝐸(𝑓), intersects the graphic of 𝑓 in one point.
a) The graphic of the function 𝑓 is represented in Fig. 2.9.
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Algebraic problems and exercises for high school
We have 𝐸(𝑓) = (−∞, +∞) and any parallel 𝑦 = 𝑚, 𝑚 ∈ ℝ to
the abscises’ axis intersects the graphic of the function 𝑓 in one point.
Consequently, 𝑓 is a bijectve function.
𝑥 2 − 6𝑥 + 8, 𝑖𝑓 𝑥 2 − 6𝑥 + 8 ≥ 0,
𝑓(𝑥) = { 2
=
−𝑥 + 6𝑥 − 8, 𝑖𝑓 𝑥 2 − 6𝑥 + 8 < 0
𝑥 2 − 6𝑥 + 8, if 𝑥 ∈ (−∞, 2] ∪ [4, +∞),
={
−𝑥 2 + 6𝑥 − 8, if 𝑥 ∈ (2,4).
Figure 2.9.
The graphic of the function 𝑓 is represented in Fig. 2.10.
We have 𝐸(𝑓) = [0; 3] ⊂ [−1; 3]. Any parallel 𝑦 = 𝑚, 𝑚 ∈
(0,1) intersects the graphic of the function 𝑓 in four points; the
parallel 𝑦 = 1 intersects the graphic in three points; any parallel 𝑦 =
𝑚, 𝑚 ∈ (1; 3] intersects the graphic of the function 𝑓 in two points.
So the function 𝑓 is neither surjective, nor injective.
c) The graphic of the function is represented in Fig. 2.11. We
have
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
Any parallel 𝑦 = 𝑚 to the abscises’ axis cuts the graphic of the
function 𝑓(𝑥) in one point, at most (𝑦 = −3, 𝑦 = 2, 𝑦 = 4) and that
is why 𝑓(𝑥) is injective. The equation 𝑓(𝑥) = 𝑟 ∈ ℝ has a solution
only for 𝑟 ≤ 3, hence 𝑓(𝑥) is not a surjective function.
d) We explain the function 𝑓:
𝑥 + 1, 𝑖𝑓 1 − 𝑥 ≤ 𝑥 + 1,
𝑓(𝑥) = max(𝑥 + 1, 1 − 𝑥) = {
=
1 − 𝑥, 𝑖𝑓 𝑥 + 1 < 1 − 𝑥
𝑥 + 1, 𝑖𝑓 𝑥 ≥ 0,
={
1 − 𝑥, 𝑖𝑓 𝑥 < 0.
Figure 2.10.
Figure 2.11.
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Algebraic problems and exercises for high school
The graphic of the function 𝑓 is represented in Fig. 2.12.
We have 𝐷(𝑓) = ℝ, 𝐸(𝑓) = [1, +∞). Any parallel 𝑦 = 𝑚 ,
𝑚 ∈ (1, +∞) intersects the graphic in two points, therefore 𝑓 is not
injective. Straight line 𝑦 = 1/2 ∈ [0, +∞) doesn’t intersect the
graphic of the function 𝑓, so 𝑓 is not surjective either.
Figure 2.12.
11. Determine which of the following relations are applications,
and which ones are injective, surjective, bijective?
For the bijective function, indicate its inverse.
Solution.
Because
the equation 2 = 𝑦 + 3 doesn’t have solutions in ℕ, it follows that 𝜑
is not a functional relation.
Then, because 𝐷(𝑓) = 𝛿𝑓 = [−1; 0], and
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
with √3/2 ≠ −√3/2, and 𝑓 is not a functional relation.
c) We have: (𝑥, 𝑦) ∈ 𝑔 ⇔ 𝑥𝑔𝑦 ⇔ 𝑦 = 𝑔(𝑥) = 𝑥 2. Then:
because 𝑥1 , 𝑥2 ∈ (0, +∞);
4) The equation 𝑔(𝑥) = 𝑟 ∈ ℝ has solutions in [0, +∞), if and
only if 𝑟 ≥ 0, which proves that 𝑔 is not a surjection (number -2, for
example, doesn’t have a preimage in [0, +∞).
d) We have:
Repeating the reasoning from c), we obtain that 𝜓 is an
injection. Furthermore, the equation 𝜓(𝑥) = 𝑟 ∈ [0, +∞) has the
solution 𝑥 = √𝑟 ∈ [0, +∞) , and that is why 𝜓 is a surjective
application. Then 𝜓 is a bijective application and, in accordance with
the 6th Theorem, the relation 𝜓 −1 is also an application (function).
In accordance with the same 6th Theorem, we have
12. Let {𝐴 = 1,2,3,4,5,6,7,8,9} and 𝜑 ∈ 𝐹(𝐴) given; using the
following table:
determine:
a)
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Algebraic problems and exercises for high school
b)
c) Calculate 𝜑 −1 (1); 𝜑 −1 (4); 𝜑 −1 (7).
Solution.
13. Let 𝐴 = {1,2,3} and 𝐵 = {𝑎, 𝑏, 𝑐}. We examine the
relations
a) Determine 𝛿𝛼 , 𝛿𝛽 , 𝛿𝛾 , and 𝜌𝛼 , 𝜌𝛽 , 𝜌𝛾 .
b) Which of the relations 𝛼, 𝛽 ș𝑖 𝛾 are applications? Indicate the
application type.
c) Determine the relations 𝛼 −1 , 𝛽 −1 and 𝛾 −1. Which one is a
function?
d) Determine the relations 𝛼 ∘ 𝛽 −1 , 𝛽 ∘ 𝛼 −1 , 𝛼 ∘ 𝛾 −1 , 𝛾 ∘ 𝛼 −1 , 𝛽 ∘
−1
𝛾 and 𝛾 ∘ 𝛽 −1. Which ones are applications?
Solution.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
b) 𝛼 is not an application, because element 3 has two images 𝑎
and 𝑐. 𝛽 is a bijective application; 𝛾 is an application, nor injective (the
elements 3 ≠ 1 have the same image c), nor surjective (b doesn’t have
a preimage).
We have:
𝛼 −1 is not an application, because the element 𝑎 has two
images 1 and 3; 𝛽 −1 is a bijective application; 𝛾 −1 is not an application,
because 𝛿𝛾−1 ≠ 𝐵 (the element 𝑏 doesn’t have an image);
Only the relation 𝛾 ∘ 𝛽 −1 is an application, neither injective, nor
surjective.
14. Given the functions: 𝑓, 𝑔: ℝ ⟶ ℝ,
3 − 𝑥, 𝑖𝑓 𝑥 ≤ 2,
𝑓(𝑥) = {
𝑔(𝑥) = max(𝑥 − 2, 3 − 𝑥),
𝑥 − 1, 𝑖𝑓 𝑥 > 2,
show that 𝑓 = 𝑔.
Solution. The functions 𝑓 and 𝑔 are defined on ℝ and have
values in ℝ. Let’s show that 𝑓(𝑥) = 𝑔(𝑥), (⩝)𝑥 ∈ ℝ.
We explain the function 𝑔:
3 − 𝑥, 𝑖𝑓 𝑥 − 1 ≤ 3 − 𝑥,
3 − 𝑥, 𝑖𝑓 𝑥 ≤ 2,
𝑔(𝑥) = {
={
= 𝑓(𝑥)
𝑥 − 1, 𝑖𝑓 3 − 𝑥 < 𝑥 − 1,
𝑥 − 1, 𝑖𝑓 𝑥 > 2,
no matter what 𝑥 ∈ ℝ is. So 𝑓 = 𝑔.
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Algebraic problems and exercises for high school
15. The functions 𝑓, 𝑔: ℝ ⟶ ℝ,
𝑥 + 2, if 𝑥 ≤ 2,
𝑥 − 2, if 𝑥 < 3,
𝑓(𝑥) = { 𝑥+10
𝑔(𝑥) = {
are given.
,
if
𝑥
>
2,
2𝑥 − 5, if 𝑥 ≥ 3
3
a) Show that 𝑓 and 𝑔 are bijective functions.
b) Represent graphically the functions 𝑓 and 𝑓 −1 in the same
coordinate register.
c) Determine the functions: 𝑠 = 𝑓 + 𝑔, 𝑑 = 𝑓 − 𝑔, 𝑝 = 𝑓. 𝑔, 𝑞 = 𝑓/𝑔.
d) Is function 𝑑 bijective?
Solution. a), b) The graphic of the function 𝑓 is represented in
Fig. 2.13 with a continuous line, and the graphic of 𝑓 −1 is represented
with a dotted line.
Figure 2.13.
We have:
𝑥 + 2, if 𝑥 < 1,
𝑥 − 2, if 𝑥 ≤ 4,
𝑓 −1 (𝑥) = {
𝑔(𝑥) = {1
3𝑥 − 10, if 𝑥 > 4,
(𝑥 + 5), if 𝑥 ≥ 1.
2
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
c) We draw the following table:
We have
2𝑥, 𝑖𝑓 𝑥 ≤ 2,
4
(𝑥
𝑠(𝑥) = 3 + 1), 𝑖𝑓 2 < 𝑥 < 3,
1
(7𝑥 − 5), 𝑖𝑓 𝑥 ≥ 3,
{ 3
4
, 𝑖𝑓 𝑥 ≤ 2,
2
2
𝑑(𝑥) =
(8 − 𝑥), 𝑖𝑓 2 < 𝑥 < 3,
3
5
(5
{ 3 − 𝑥), 𝑖𝑓 𝑥 ≥ 3,
𝑥 2 − 4, 𝑖𝑓 𝑥 ≤ 2,
1 2
𝑝(𝑥) = 3 (𝑥 + 8𝑥 − 20), 𝑖𝑓 2 < 𝑥 < 3,
1
(2𝑥 2 + 15𝑥 − 50), 𝑖𝑓 𝑥 ≥ 3,
{3
𝑥+2
, 𝑖𝑓 𝑥 ≤ 2,
𝑥−2
𝑥 + 10
, 𝑖𝑓 2 < 𝑥 < 3,
𝑞(𝑥) =
3(𝑥 − 2)
𝑥 + 10
, 𝑖𝑓 𝑥 ≥ 3.
{ 3(2𝑥 − 5)
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Algebraic problems and exercises for high school
d) We have 𝑑(𝑥) = 4, (⩝)𝑥 ∈ (−∞, 2], therefore 𝑑 is not
injective, in other words, 𝑑 is not bijective. This proves that the
difference (sum) of two bijective functions isn’t necessarily a bijective
function.
16. Let’s consider the function 𝑓: ℝ ∖ {−𝑑/𝑐} ⟶ ℝ ∖ {𝑎/𝑐},
a) Show that function 𝑓 is irreversible.
b) Determine 𝑓 −1.
c) In what case do we have 𝑓 = 𝑓 −1?
Solution:
⇔ (𝑎𝑑 − 𝑏𝑐)(𝑥1 − 𝑥2 ) = 0 ⇔ 𝑥1 = 𝑥2 ⇒ 𝑓 is injective.
2) Let 𝑥 ∈ ℝ ∖ {𝑎/𝑐}. We examine the equation 𝑓(𝑥) = 𝑟. Then
If
Imposibile. So (𝑑𝑟 − 𝑏)/(𝑎 − 𝑐𝑟) ∈ ℝ ∖ {−𝑑/𝑐}, which proves
that the equation 𝑓(𝑥) = 𝑟 has solutions in ℝ ∖ {−𝑑/𝑐} for any ℝ ∖
{𝑎/𝑐}. This proves that 𝑓 is a surjective function.
From 1) - 2), it follows that 𝑓 is a bijective function, hence it is
also inversibile.
b) We determine
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
c) In order to have 𝑓 = 𝑓 −1, it is necessary that the functions 𝑓
and 𝑓 −1 have the same definition domain and so 𝑑 = −𝑎. This
condition is also sufficient for the equality of the functions 𝑓 and 𝑓 −1 .
It is true that, if 𝑑 = −𝑎, the functions 𝑓 and 𝑓 −1are defined on ℝ ∖
{𝑎/𝑐}, take values in ℝ ∖ {𝑎/𝑐}, and 𝑓 −1 (𝑥) =
𝑎𝑥+𝑏
𝑐𝑥+𝑑
= 𝑓(𝑥), (⩝)𝑥 ∈
𝑎
ℝ ∖ { }, namely 𝑓 = 𝑓 −1.
𝑐
17. Represent the function 𝑓(𝑥) = √5𝑥 2 − 2𝑥 + 8 as a
compound of two functions.
Solution. We put
Then
Answer:
18. Calculate 𝑓 ∘ 𝑔, 𝑔 ∘ 𝑓, (𝑓 ∘ 𝑔)(4) and (𝑔 ∘ 𝑓)(4), where:
3
a) 𝑓(𝑥) = 𝑥−1 and 𝑔(𝑥) = √𝑥;
b) 𝑓(𝑥) = √𝑥 + 3 and 𝑔(𝑥) = 𝑥 2 − 4;
𝑥 2 + 6𝑥, 𝑖𝑓 x ≤ −3
c) 𝑓(𝑥) = {
−2𝑥 − 5, 𝑖𝑓 𝑥 ≥ −3
5𝑥 − 2, 𝑖𝑓 𝑥 ≤ 1,
and 𝑔(𝑥) = { 2
𝑥 − 2𝑥 + 4, 𝑖𝑓 𝑥 > 1.
Indicate 𝐷(𝑓 ∘ 𝑔) and 𝐷(𝑔 ∘ 𝑓).
Solution: a) We have
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Algebraic problems and exercises for high school
So (𝑔 ∘ 𝑓)(4) ≠ (𝑓 ∘ 𝑔)(4), which proves that the comutative
law of function composition, generally, does not take place: 𝑓 ∘ 𝑔 ≠
𝑔 ∘ 𝑓.
c) To determine the functions 𝑓 ∘ 𝑔 and 𝑔 ∘ 𝑓, in this particular
case we proceed as it follows:
𝑔2 (𝑥) + 6𝑔(𝑥),
𝑖𝑓 𝑔(𝑥) < −3
(𝑓 ∘ 𝑔)(𝑥) = (𝑓 ∘ 𝑔)(𝑔(𝑥)) == {
=
−2𝑔(𝑥) − 5,
𝑖𝑓 𝑔(𝑥) ≥ −3
𝑔2(𝑥) + 6𝑔(𝑥) 𝑖𝑓 𝑔(𝑥) < −3,
−2𝑔(𝑥) − 5, 𝑖𝑓 𝑔(𝑥) ≥ −3 and 𝑥 ≤ 1,
={
=
−2(𝑥 2 − 2𝑥 + 4) − 5, 𝑖𝑓 (𝑥) ≥ −3 and 𝑥 > 1
(5𝑥 − 2)2 + 6(5𝑥 − 2) 𝑖𝑓 5𝑥 − 2 < −3,
{ −2(5𝑥 − 2) − 5, 𝑖𝑓 5𝑥 − 2 ≥ −3 and 𝑥 ≤ 1, =
−2(𝑥 2 − 2𝑥 + 4) − 5, 𝑖𝑓 5𝑥 − 2 ≥ −3 and 𝑥 > 1
25𝑥 2 + 10𝑥 − 8 𝑖𝑓 𝑥 < −1/5,
= {−10𝑥 − 1, 𝑖𝑓 − 1/5 ≤ 𝑥 ≤ 1,
2𝑥 2 + 4𝑥 − 13, 𝑖𝑓 𝑥 > 1.
5𝑓(𝑥),
𝑖𝑓 𝑓(𝑥) ≤ 1,
(𝑔 ∘ 𝑓)(𝑥) = 𝑔(𝑓(𝑥)) = { 2
𝑓 (𝑥) − 𝑠𝑓(𝑥) + 4,
𝑖𝑓 𝑓(𝑥) > 1.
We notice that
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
So for 𝑓(𝑥) ≤ 1, we obtain
5(𝑥 2 + 6𝑥) − 2,
if 𝑥 ∈ [−3 − √10, −3)
(𝑔 ∘ 𝑓)(𝑥) = {
5(−2𝑥 − 5) − 2,
if 𝑥 ∈ [−3, +∞).
Similarly,
and that is why for 𝑓(𝑥) > 1, we obtain
for 𝑥 ∈ (−∞, −3 − √10).
To sum it up, we have
(𝑔 ∘ 𝑓)(𝑥) =
𝑥 4 + 12𝑥 3 + 34𝑥 2 − 12𝑥 + 4,
𝑖𝑓 𝑥 ∈ (−∞, −3 − √10),
2
={
5𝑥 + 30𝑥 − 2,
if 𝑥 ∈ [−3√10, −3),
−10𝑥 − 7, if 𝑥 ≥ −3.
(𝑔 ∘ 𝑓)(4) = −10.4 − 27 = −67.
19. Solve the following equations:
a) (𝑔 ∘ 𝑓 ∘ 𝑓)(𝑥) = (𝑓 ∘ 𝑔 ∘ 𝑔)(𝑥) , if 𝑓(𝑥) = 3𝑥 + 1, 𝑔(𝑥) =
𝑥 + 3;
b) (𝑓 ∘ 𝑓 ∘ 𝑓)(𝑥) = 𝑥, if 𝑓(𝑥) =
𝑎𝑥+1
𝑥+𝑎
, 𝑎 ∈ ℝ, 𝑥 ∈ ℝ ∖ {−𝑎};
c) (𝑓 ∘ 𝑔)(𝑥) = (𝑔 ∘ 𝑓)(𝑥), if 𝑓(𝑥) = 2𝑥 2 − 1, 𝑔(𝑥) = 4𝑥 3 − 3𝑥.
Solution. We determine the functions (𝑔 ∘ 𝑓 ∘ 𝑓)(𝑥) and (𝑓 ∘
𝑔 ∘ 𝑔)(𝑥):
The equation becomes
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Algebraic problems and exercises for high school
Answer: 𝑥 = 2.
b) We calculate (𝑓 ∘ 𝑓 ∘ 𝑓)(𝑥):
The equation becomes
Answer: 𝑥 ∈ {−1,1}
c) We determine (𝑓 ∘ 𝑔) and (𝑔 ∘ 𝑓)(𝑥):
The equation becomes
meaning the equality is true for any 𝑥 ∈ ℝ.
Answer: 𝑥 ∈ ℝ.
20. Let 𝑓, 𝑔: ℝ ⟶ ℝ given by 𝑓(𝑥) = 𝑥 2 + 𝑥 + 12 and 𝑔(𝑥) =
𝑥 2 − 𝑥 + 2. Show there is no function 𝜑: ℝ ⟶ ℝ, so that
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
Solution. The relation (𝐴) can also be written
We suppose there is a function 𝜑: ℝ ⟶ ℝ that satisfies the
relation (𝐴′). We put in (𝐴′) 𝑥 = 1 and 𝑥 = −1. We obtain:
which implies 𝜑(4) +
𝜑(2) ≠ 𝜑(2) + 𝜑(4), contradiction with the hypothesis. So there is
no function 𝜑 with the property from the enunciation.
21. Determine all the values of the parameters 𝑎, 𝑏 ∈ ℝ for
which (𝑓 ∘ 𝑔) = (𝑔 ∘ 𝑓)(𝑥) , (⩝) ∈ ℝ , where 𝑓(𝑥) = 𝑥 2 − 𝑥 and
𝑔(𝑥) = 𝑥 2 + 𝑎𝑥 + 𝑏.
Solution. We determine 𝑓 ∘ 𝑔 and 𝑔 ∘ 𝑓:
Then
Answer: 𝑎 = −1, 𝑏 = 0, 𝑔(𝑥) = 𝑥 2 − 𝑥 = 𝑓(𝑥).
22. Given the functions 𝑓, 𝑔, ℎ ∶ ℝ ⟶ ℝ,
.
Show that:
a) 𝑓 is not injective;
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Algebraic problems and exercises for high school
b) 𝑔 is injective;
c) ℎ is bijective and determine ℎ−1.
Solution. a) Let 𝑓(𝑥1 ) = 𝑓(𝑥2 ).
Then
For example, 𝑓(𝑥) = 𝑥 3 (𝑥 + 4) + 3, taking 𝑥1 = 0 and 𝑥2 = −4,
we obtain 𝑓(0) = 𝑓(−4) = 3, 𝑥1 ≠ 𝑥2.
because
so ℎ is an injective function.
We prove that ℎ is surjective. Let 𝑟 ∈ ℝ. We solve the equation
(the root of an uneven order exists out of any real number). So ℎ is a
surjection, therefore, ℎ is a bijective function.
We determine ℎ−1:
23. Let 𝑓: [1, +∞) ⟶ [1, +∞) with
a) Prove that 𝑓 is a bijective function.
b) Determine 𝑓 −1.
Solution. a) We have 𝑓(𝑥) = (𝑥 2 − 𝑥 + 1) 3 . We represent this
function as a compound of two functions:
𝑢, 𝑣: [1, +∞) ⟶ [1, +∞), where 𝑢(𝑥) = 𝑥 3, 𝑣(𝑥) = 𝑥 2 − 𝑥 + 1.
Then 𝑓(𝑥) = (𝑢 ∘ 𝑣)(𝑥) , which implies that 𝑓 = 𝑢 ∘ 𝑣 is bijective,
being the compound of two bijective functions.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
b) 𝑓 −1 = (𝑢 ∘ 𝑣)−1 = 𝑣 −1 ∘ 𝑢−1. We determine 𝑣 −1 and 𝑢−1 .
Because:
We solve this equation according to 𝑥 ∈ [1, +∞).
The discriminant is:
The equation has only one root in [1, +∞):
Then
So,
3
= (1 + √4 √𝑥 − 3)/2, with 𝑓 −1 : [1, +∞) ⟶ [1, +∞).
24. Considering the function 𝑓: ℝ ⟶ ℝ with the properties:
a) Determine the function 𝑓.
b) Calculate 𝑓(√1998).
Solution. a) For 𝑥2 = 0 from 1), it follows that 𝑓(𝑥1 ) = 𝑓(𝑥1 ) +
𝑓(0), which implies that 𝑓(0) = 0. For 𝑥2 = −𝑥1 from 1), we obtain
Then
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Algebraic problems and exercises for high school
Let 𝑥 ∉ {0, 1}. Then
On the other hand,
1
1−𝑥
=
1−𝑥+𝑥
1−𝑥
=1+
𝑥
1−𝑥
implies
From (2) and (3) it follows that
So 𝑓(𝑥) = 𝑥, (⩝) ∈ ℝ.
Answer: a) 𝑓(𝑥) = 𝑥; b) 𝑓(√1998) = √1998.
25. Using the properties of the characteristic function, prove the
equality:
Solution. Using the properties 𝐴 = 𝐵 ⇔ 𝑓𝐴 = 𝑓𝐵 , we will
prove the required equality by calculating with the help of 𝑓𝐴 , 𝑓𝐵 and
𝑓𝐶 the characteristic functions of sets 𝐴 ∪ (𝐵 ∩ 𝐶) and (𝐴 ∪ 𝐵) ∩
(𝐴 ∪ 𝐶):
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
which implies 𝐴 ∪ (𝐵 ∩ 𝐶) and (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶).
2.3. Suggested exercises
1. Determine the definition domains and the values domains of
the following relations:
2. Let 𝐴 = {2,4,6,8} and 𝐵 = {1,3,5,7}, 𝛼 ⊆ 𝐴 × 𝐵.
a) Determine the graphic of the relation 𝛼.
b) Construct the schema of the relation 𝛼:
1) 𝛼 = {(𝑥, 𝑦)│𝑥 < 3 and 𝑦 > 3};
2) 𝛼 = {(𝑥, 𝑦)│𝑥 > 2 and 𝑦 < 5};
3) 𝛼 = {(𝑥, 𝑦)│𝑥 > 6 or 𝑦 > 7};
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Algebraic problems and exercises for high school
3. Let 𝐴 = {1,2,3,4} and 𝐵 = {1,3,5,7,8}. Write the graphic of
the relation 𝛼 ⊆ 𝐴 × 𝐵, if:
4. Let 𝐴 = {𝑙, 3, 4, 5}, 𝐵 = {𝑙, 2, 5, 6} and 𝐺 be the graphic of
the relation 𝛼. Write the relation 𝛼 with propositions containing letters
𝑥 and 𝑦, with 𝑥 𝐸 𝐴 and 𝑦 ∈ 𝐵:
5. Let A = {𝑙, 2.3,4}. Analyze the properties of the relation 𝛼 ⊆
𝐴 (var. 1-6) and 𝛼 ⊆ ℝ2 (var. 7-14):
2
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
7) 𝛼 = {(𝑥, 𝑦) ∈ ℝ2 │𝑥 > 1 and 𝑦 > 1 };
𝑥 > 0,
𝑥 < 0,
8) 𝛼 = {(𝑥, 𝑦) ∈ ℝ2 │ {
or {
};
𝑦>0
𝑦<0
9) 𝛼 = {(𝑥, 𝑦) ∈ ℝ2 │𝑥 ≥ 0 or 𝑦 ≥ 0 };
10) 𝛼 = {(𝑥, 𝑦) ∈ ℝ2 │𝑥 ≥ 1 or 𝑦 ≥ 1 };
6. For each binary relation 𝛼 defined on set ℕ:
a) determine the definition domain 𝛿𝛼 and the values domain 𝜌𝛼 ;
b) establish the properties (reflexivity, ireflexivity, symmetry,
anti-symmetry, transitivity);
c) determine the inverse relation 𝛼 −1 (𝑥, 𝑦 ∈ ℕ):
7. Given the set 𝐴 and the binary relation ⊆ 𝐴2 , prove that 𝛼 is
an equivalence relation and determine the factor set 𝐴/𝛼.
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Algebraic problems and exercises for high school
8. Given the set 𝐴 = {1, 2,3,4,5,6,7,8, 9} and the subset system
𝑆 = {𝐴𝑖 ⊆ 𝐴, 𝐼 = ̅̅̅̅̅
1, 𝑛}, prove that 𝑆 defines a partition on 𝐴 and build
the equivalence relation 𝛼𝑆 .
9. We define on ℝ the binary relation 𝛼:
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
a) Show that 𝛼 is an equivalence relation on ℝ.
b) Determine the equivalence classes.
10. We define on ℝ the binary relation 𝛽:
.
a) Show that 𝛽 is an equivalence relation.
b) Determine the equivalence classes.
11. Let ⊆ 𝐴2 , where 𝐴 = {−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, 6,
7, 8} with
a) Is 𝛼 an equivalence relation?
b) If so, determine the equivalence classes.
12. Determine: 𝛿𝛼 , 𝜌𝛼 , 𝛼 −1 , 𝛼 ∘ 𝛼, 𝛼 ∘ 𝛼 −1 , 𝛼 −1 ∘ 𝛼, if:
13. Determine the relations
𝛼 ∘ 𝛽, 𝛽 ∘ 𝛼, 𝛼 −1 , 𝛽 −1, 𝛼 −1 ∘ 𝛽, (𝛽 ∘ 𝛼)−1 ;
5) 𝛼 = {(𝑥, 𝑦) ∈ ℤ2 │𝑥 − 𝑦 𝑖s even},
𝛽 = {(𝑥, 𝑦) ∈ ℤ2 │𝑥 − 𝑦 is uneven};
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Algebraic problems and exercises for high school
14. Let 𝐴 = {1, 2, 3, 4}, 𝐵 = {𝑎, 𝑏, 𝑒, 𝑑} . Determine the
definition domain and the values domain of each of the following
relations 𝛼, 𝛽. Which ones are applications? Determine the application
type. Determine the relation 𝛼 −1 ∘ 𝛼, 𝛼 ∘ 𝛽 −1, 𝛽 ∘ 𝛼 −1, 𝛽 ∘ 𝛽 −1. Are
they applications?
15. We consider the application 𝜑: ℝ ⟶ ℝ , 𝜑(𝑥) = sin 𝑥 .
Determine:
16. The application 𝜑: 𝐴 ⟶ 𝐵 is given by the table below, where
𝐴 = {1, 2, 3, 4, 5, 6, 7, 8, 9} and 𝐵 = {𝑎, 𝑏, 𝑒, 𝑑, 𝑒, 𝑖}.
Determine:
17. We consider the application 𝜑: ℝ ⟶ ℤ, 𝜑( 𝑥) = [𝑥] ([𝑥] is
the integer part of 𝑥). Determine
𝜑 −1 ({2,4,5}) and 𝜑 −1(−1).
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
18. The following application is given: 𝜑: ℕ ⟶ ℕ, 𝜑(𝑥) = 𝑥 2.
Determine 𝜑(𝐴) and 𝜑 −1 (𝐴), if 𝐴 = {1, 2, 3, 4, 5, 6 , 7, 8, 9, 10}.
19. Let 𝐴 and 𝐸 be two finite sets, |𝐴| = 𝑚. |𝐵|. How many
surjective applications 𝜑: 𝐴 ⟶ 𝐵 are there, if:
20. Given the graphic of the relation 𝛼, establish if 𝛼 is a function. Determine 𝛿𝛼 and 𝜌𝛼 .
Changing 𝛿𝛼 and 𝜌𝛼 , make 𝛼 become an application injective,
surjective and bijective.
Draw the graphic for the relation 𝛼 −1. Is relation 𝛼 −1 a function?
What type is it?
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Algebraic problems and exercises for high school
21. Let 𝐴 = {1, 2, 3, 4}, 𝐵 = {0, 1, 5, 6} and 𝐶 = {7, 8, 9}.
a) Draw the diagrams of two injective and two non-injective
functions defined on 𝐶 with values in 𝐵.
b) Draw the diagrams of two surjective functions and two nonsurjective applications defined on 𝐴 with values in 𝐶.
c) Draw the diagrams of two bijective and two non-bijective
functions defined on 𝐴 with values in 𝐵.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
22. Let 𝐴 = {1, 3, 5, 6} and 𝐵 = {0, 1, 2, 3, 5}, 𝑥 ∈ 𝐴 , 𝑦 ∈ 𝐵 .
Which of the relations stated below represents a function defined on 𝐴
with values in 𝐵? And one defined on 𝐵 with values in 𝐴? For functions,
indicate their type:
23. Using the graphic of the function 𝑓: 𝐴 ⟶ 𝐵, show that 𝑓 is
injective, surjective or bijective. If bijective, determine 𝑓 −1 and draw
the graphic of 𝑓 and 𝑓 −1in the same register of coordinates..
24. Which of the following relations 𝑎 ⊆ ℝ2 are functions?
Indicate the definition domain of the functions. Establish their type:
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Algebraic problems and exercises for high school
25. Given the relation 𝛼 with 𝛽𝛼 = [−3; 5] and 𝜌𝛼 = [−4; 7]:
a) Does pair (−4, 5) belong to the relation 𝛼? Why?
b) Indicate all the ordered pairs (𝑥, 𝑦) ∈ 𝛼 with 𝑥 = 𝑂. Explain.
26. Given the function 𝑓(𝑥), calculate its values in the indicated
points.
𝑥 2 − 5, 𝑖𝑓 𝑥 > 0,
4) 𝑓(𝑥) = {
𝑓(−2), 𝑓(0), 𝑓(5);
2𝑥 + 3, 𝑖𝑓 𝑥 ≤ 0,
𝑥 2 + 1, 𝑖𝑓 𝑥 > 0
5) 𝑓(𝑥) = { −4, 𝑖𝑓 𝑥 = 0 𝑓(5), 𝑓(−1), 𝑓(1/2);
1 − 2𝑥, 𝑖𝑓 𝑥 < 0
27. Determine 𝐷(𝑓), if:
28. Given the functions 𝑓(𝑥) and 𝑔(𝑥), determine the functions
𝑓 + 𝑔, 𝑓 − 𝑔, 𝑓 · 𝑔 and 𝑓 / 𝑔, indicating their definition domain.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
29. Represent the function 𝑓(𝑥) as a compound of some
functions:
30. Given the functions 𝑓 (𝑥) and 𝑔( 𝑥 ), determine the
functions 𝑓 ∘ 𝑔 and 𝑔 ∘ 𝑓 and calculate (𝑓 ∘ 𝑔 )(3) and (𝑔 ∘ 𝑓 )(3) in
options 1) − 6) and (𝑓 ∘ 𝑔 )( −1) and (𝑔 ∘ 𝑓)( -1) in options 7) −
12).
31. Given the functions 𝑓(𝑥) = 𝑥 2 , 𝑔(𝑥) = 3𝑥 and ℎ(𝑥) = 𝑥 −
1, calculate:
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Algebraic problems and exercises for high school
32. Determine if in the function pairs 𝑓 and 𝑔 one is the inverse
of the other:
33. The value of function 𝑓 is given. Calculate the value of the
function 𝑓 −1, if:
34. Determine 𝑓 −1, if:
35. The function: 𝑓: ∈ ℝ ⟶ ℝ is given:
a) Prove that 𝑓 is bijective.
b) Determine 𝑓 −1
c) Calculate 𝑓 ∘ 𝑓 −1 ș𝑖 𝑓 −1 ∘ 𝑓.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
36. Given the functions 𝑓( 𝑥) and 𝑔( 𝑥), determine the functions
(𝑓 ∘ 𝑔)(𝑥) and (𝑔 ∘ 𝑓)(𝑥)(𝑓, 𝑔: ℝ ⟶ ℝ):
37. Prove the equality of the functions 𝑓 and 𝑔:
38. Determine the functions 𝑠 = 𝑓 + 𝑔, 𝑑 = 𝑓 − 𝑔, 𝑝 = 𝑓. 𝑔
and 𝑞 = 𝑓/𝑔:
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Algebraic problems and exercises for high school
39. Let 𝐴 = {1, 2, 3, 4}, 𝐵 = {0, 1, 3, 4} and the functions
Can the functions 𝑓 ∘ 𝑔, 𝑔 ∘ 𝑓 be defined? If so, determine
these functions. Make their diagrams.
40. a) Show that the function 𝑓 is bijective.
b) Determine 𝑓 −1.
c) Graphically represent the functions 𝑓 and 𝑓 −1 in the same
coordinate register.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
41. Show that the function 𝑓: ∈ ℝ ⟶ ℝ, 𝑓(𝑥) = 𝑥 2 − 6𝑥 + 2
admits irreversible restrictions on:
Determine the inverses of these functions and graphically
represent them in the same coordinate register.
42. Using the properties of the characteristic function, prove the
following equalities (affirmations):
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Algebraic problems and exercises for high school
3. Elements of combinatorics
3.1. Permutations. Arrangements.
Combinations. Newton’s Binomial
In solving many practical problems (and beyond), it is necessary
to:
1) be able to assess the number of different combinations
compound with the elements of a set or a multitude of sets;
2) choose (select) from a set of objects the subsets of elements
that possess certain qualities;
3) arrange the elements of a set or a multitude of sets in a
particular order etc.
The mathematical domain that deals with these kind of
problems and with the corresponding solving methods is called
combinatorics. In other words, combinatorics studies certain
operations with finite sets. These operations lead to notions of
permutations, arrangements and combinations.
Let 𝑀 = {𝑎1 , 𝑎2 , … , 𝑎𝑛 } be a finite set that has 𝑛 elements. Set
𝑀 is called ordered, if each of its elements associates with a certain
number from 1 to 𝑛, named the element rank, so that different
elements of 𝑀 associate with different numbers.
Definition 1. All ordered sets that can be formed with 𝑛 elements
of given set 𝑀 (𝑛( 𝑀) = 𝑛) are called permutations of 𝑛 elements.
The number of all permutations of 𝑛 elements is denoted by the
symbol 𝑃𝑛 and it is calculated according to the formula:
By definition, it is considered 𝑃0 = 0! = 1 = 1! = 𝑃1.
Definition 2. All ordered subsets that contain 𝑚 elements of set
𝑀 with 𝑛 elements are called arrangements of 𝑛 elements taken 𝑚 at
a time.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
The number of all arrangements of 𝑛 elements taken 𝑚 at a
time is denoted by the symbol 𝐴𝑚
𝑛 and it is calculated with the formula
Definition 3. All subsets that contain elements of set 𝑀 with n
elements are called combinations of 𝑛 elements taken 𝑚 at a time.
The number of all combinations of 𝑛 elements taken 𝑚 at a time
is denoted by the symbol 𝐶𝑛𝑚 and it is calculated with the formula
where 𝑚, 𝑛 ∈ ℕ; 0 ≤ 𝑚 ≤ 𝑛.
Remark. In all the subsets mentioned in definitions 1 - 3, each
element of the initial set 𝑀 appears only once.
Along with the combinations in which each of the different 𝑛
elements of a set participates only once, we can also consider
combinations with repetitions, that is, combinations in which the same
element can participate more than once.
Let 𝑛 groups of elements be given. Each group contains certain
elements of the same kind.
Definition 1′. Permutations of 𝑛 elements each one containing
𝛼1 elements 𝑎𝑖1 , 𝛼2 elements 𝑎𝑖2 , 𝛼𝑘 elements 𝑎𝑖𝑘 , where 𝛼1 + 𝛼2 +
⋯ + 𝛼𝑘 are called permutations of 𝑛 elements with repetitions.
The number of all permutations with repetitions is denoted by
the symbol 𝑃̅𝛼1 ,𝛼2 ,…,𝛼𝑘 and it is calculated using the formula:
Definition 2′ . The arrangements of 𝑛 elements, each one
containing 𝑚 elements, and each one being capable of repeating itself
in each arrangement for an arbitrary number of times, but not more
than 𝑚 times, are called arrangements of 𝑛 elements taken 𝑚 at a
time with repetitions.
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Algebraic problems and exercises for high school
The number of all arrangements with repetitions of 𝑛 elements
taken 𝑚 at a time is denoted by the symbol ̅̅̅̅
𝐴𝑚
𝑛 and it is calculated with
the formula
Definition 3'. The combinations of 𝑛 elements each one
containing 𝑚 elements, and each one being capable of repeating itself
more than once, but not more than 𝑚 times, are called combinations
of 𝑛 elements taken 𝑚 at a time with repetitions.
The number of all combinations with repetitions is denoted by
𝑚
̅̅̅̅
the symbol 𝐶
𝑛 and it is calculated with the formula
In the process of solving combinatorics problems, it is
important to firstly establish the type (the form) of combination. One
of the rules used to establish the type of combination could be the
following table
Caution is advised to the order the elements are arranged in:
If the order isn’t important,
we have combinations;
Arrangements if not all of
the elements from the
initial set participate;
If the order is important,
we have arrangements
or permutations:
Permutations if all
elements of the initial
set participate;
It is often useful to use the following two rules:
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
The sum rule. If the object 𝐴 can be chosen in 𝑚 ways and the
object 𝐵 in n ways, then the choice "either A or B" can be made in 𝑚 +
𝑛 ways.
The multiplication rule. If the object 𝐴 can be chosen in 𝑚
ways and after each such choice the object 𝐵 can be chosen in 𝑛 ways,
then the choice "A and B" in this order can be made in 𝑚 . 𝑛 ways.
We will also mention some properties of combinations, namely:
The formula
is called Newton’s Binomial formula (𝑛 ∈ ℕ∗ ).
The coefficients 𝐶𝑛0 , 𝐶𝑛1 , 𝐶𝑛2 , … , 𝐶𝑛𝑛 from Newton’s Binomial
formula are called binomial coefficients; they possess the following
qualities:
V. The binomial coefficients from development (7), equally set
apart from the extreme terms of the development, are equal among
themselves.
VI a. The sum of all binomial coefficients equals 2𝑛 .
VI b. The sum of the binomial coefficients that are on even
places equals the sum of the binomial coefficients that are on uneven
places.
VII. If 𝑛 is an even number (i.e. 𝑛 = 2𝑘), then the binomial
coefficient of the middle term in the development (namely 𝐶𝑛𝑘 ) is the
biggest. If 𝑛 is uneven (i.e. 𝑛 = 2𝑘 + 1 ), then the binomial
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Algebraic problems and exercises for high school
coefficients of the two middle terms are equals (namely 𝐶𝑛𝑘 = 𝐶𝑛𝑘+1)
and are the biggest.
VIII. Term 𝐶𝑛𝑘 𝑥 𝑛−𝑘 𝑎𝑘 , namely the (𝑘 + 1) term in equality (7),
is called term of rank 𝑘 + 1 (general term of development) and it is
denoted by 𝑇𝑘+1 . So,
IX. The coefficient of the term of rank 𝑘 + 1 in the
development of Newton’s binomial is equal to the product of the
coefficient term of rank 𝑘 multiplied with the exponent of 𝑥 in this
term and divided by 𝑘, namely
3.2. Solved problems
1. In how many ways can four books be arranged on a shelf?
Solution. As order is important and because all the elements of
the given set participate, we are dealing with permutations.
So
Answer: 24.
2. A passenger train has ten wagons. In how many ways can the
wagons be arranged to form the train?
Solution. As in the first problem, we have permutations of 10
elements of a set that has 10 elements. Then the number of ways the
train can be arranged is
Answer: 3 628 800.
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3. In how many ways can 7 students be placed in 7 desks so that
all the desks are occupied?
Solution.
Answer: 5 040.
4. How many phone numbers of six digits can be dialed:
1) the digit participates in the phone number only once;
2) the digit participates more than once?
(The telephone number can also start with 0.)
Solution. We have 10 digits on the whole: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
As the phone number can also start with 0, we have:
1) arrangements of 10 digits taken 6 at a time, i.e.
2) because the digit can repeat in the number, we have
arrangements with repetitions, i.e.
Answer: 1) 151 200; 2) 1 000 000.
5. The volleyball team is made out of 6 athletes. How many
volleyball teams can a coach make having 10 athletes at his disposal?
Solution. As the coach is only interested in the team’s
composition it is sufficient to determine the number of combinations
of 10 elements taken 6 at a time, i.e.
Answer: 210.
6. How many 5 digit numbers can be formed with digits 0 and 1?
Solution. As the digits are repeating and their order is important,
we consequently have arrangements with repetitions. Here, 𝑚 = 5,
𝑛 = 2.
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Algebraic problems and exercises for high school
From digits 0 and 1 we can form ̅̅̅
𝐴52 numbers of five digits.
But we have to take into consideration that the number cannot
begin with the digit zero. So, from the number ̅̅̅
𝐴52 we have to begin
with zero. Numbers of this type are ̅̅̅
𝐴4 . Therefore, the number we
2
seek is
Answer: 16.
7. How many three digit numbers can be formed with the digits
1, 2, 3, 4, 5, if the digits can be repeated?
Solution. As the digits repeat, we obviously have arrangements
with 5 digits taken three times. Therefore, there can be formed
̅̅̅̅̅̅̅̅̅̅
𝐴3 = 53 numbers of three digits.
5
Answer: 125.
8. Using 10 roses and 8 Bedding Dahlias, bunches are made that
contain 2 roses and 3 Bedding Dahlias. How many bunches of this kind
can be formed?
Solution. Two roses (out of the 10 we have) can be chosen in
2
𝐶10 ways, and three Bedding Dahlias (out of 8) can be taken in 𝐶83 ways.
Applying the rule of multiplication, we have: the total number of
2
bunches that can be formed is 𝐶10
. 𝐶83 = 1 890.
Answer: 1890.
9. 12 young ladies and 15 young gentlemen participate at a
dancing soirée. In how many ways can four dancing pairs be chosen?
Solution. The 12 young ladies can be distributed in four person
4
4
groups in 𝐶12
ways, and the 15 young gentlemen - in 𝐶15
ways.
Because in each group of young ladies (or young gentlemen)
order is an issue, each of these groups can be ordered in 𝑃4ways.
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As a result (we apply the multiplication rule), we will have
4
4
4
4
= 𝐴12
. 𝐶15
= 𝐶12
. 𝐴15
= 16216200.
Answer: 16216200.
4
4
𝐶12
. 𝑃4 . 𝐶15
10. To make a cosmic flight to Mars, it is necessary to constitute
the crew of the spacecraft in the following distribution: the spacecraft
captain, the first deputy of the captain, the second deputy of the
captain, two mechanical engineers, and a doctor. The command triplet
can be selected out of the 25 pilots that are flight-ready, two
mechanical engineers out of the 20 specialists who master the
construction of the spacecraft, and the doctor – out of the 8 available
doctors. In how many ways can the crew be assembled?
Solution. Choosing the captain and his deputies, it is important
to establish which of the pilots would best deal with random steering
commands. So, the distribution of the tasks between the triplet’s
members is equally important. So, the commanding triplet can be
formed in 𝐴325 ways.
The tasks of the two engineers are more or less the same. They
can accomplish these tasks consecutively. So the pair of engineers can
2
be formed in 𝐶20
ways. Regarding the doctor – the situation is the
same, namely the doctor can be chosen in 𝐶81 ways.
Using the multiplication rule, we have: a pair of engineers can be
2
assigned in 𝐶20
ways to each commanding triplet. We will have, in
3
2
total, 𝐴25 . 𝐶20 quintets. To each quintet a doctor can be associated in
𝐶81 ways.
As a result, the spacecraft’s crew can be assembled in
3
2
𝐴25 . 𝐶20 . 𝐶81 ways, or
Answer: 20 976 000.
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Algebraic problems and exercises for high school
11. In how many different ways can 5 cakes, of the same type or
different, be chosen in a cake shop where there are 11 types of
different cakes?
Solution. The five cakes can all be of the same type or four of the
same type and one of a different type, or three of the same type and
two of a different type, etc., or all can be of different types.
The number of the possible sets comprised of five cakes out of
the existing 11 types is equal to the number of the combinations with
repetitions of 11 elements taken five at a time, i.e.
Answer: 3003.
12. In a group of 10 sportsmen there are two rowers, three
swimmers and the others are athletes. A 6 person team must be
formed for the upcoming competitions in such a manner that the team
will comprise at least one representative from the three nominated
sport types. In how many ways can such a team be assembled?
Solution. a) In the team there can be one rower, one swimmer
and four athletes. The rower can be chosen in 𝐶21 ways, the swimmer in
𝐶31 ways, and the athlete in 𝐶54 ways. Using the multiplication rule, we
have 𝐶21 . 𝐶31 . 𝐶54 ways.
b) In the team there can be one rower, two swimmers and three
athletes. According to the aforementioned reasoning, the numbers of
the teams of this type will be 𝐶21 . 𝐶31 . 𝐶54 .
c) In the team there can be one rower, three swimmers and two
athletes. The number of teams in this case will be 𝐶21 . 𝐶33 . 𝐶52.
d) In the team there can be two rowers, one swimmer and three
athletes. We will have 𝐶22 . 𝐶31 . 𝐶53 of these teams.
e) In the team there can be two rowers, two swimmers and two
athletes. The number of teams will be 𝐶22 . 𝐶32 . 𝐶52.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
f) In the team there can be two rowers, three swimmers and one
athlete. The number of teams will be 𝐶22 . 𝐶33 . 𝐶51.
Using the addition rule, the total number of teams that can be
assembled is:
Answer: 175.
13. Given 𝑘 = 15 capital letters, 𝑚 = 10 vowels and 𝑛 = 11
consonants (in total 𝑘 + 𝑚 + 𝑛 = 36 letters) determine: How many
different words can be formed using these letters, if in each word the
first letter has to be a capital letter, among the other different letters
there have to be 𝜇 = 4 different vowels (out of the 𝑚 = 10 given)
and 𝑣 = 6 different consonants (out of the 𝑛 = 11 given).
Solution. We choose a capital letter. This choice can be made in
𝑘 ways. Then from 𝑚 vowels we choose 𝜇 letters. This can be done in
𝜇
𝐶𝑚 ways. Finally, we choose 𝑣 consonants, which can be made in 𝐶𝑛𝑣
ways. Using the multiplication rule, we can choose the required letters
𝜇
to form the word in 𝑘. 𝐶𝑚 . 𝐶𝑛𝑣 ways.
After placing the capital letter at the beginning, with the other
𝜇 + 𝑣 letters we can form (𝜇 + 𝑣)! permutations. Each such
𝜇
permutation yields a new word. So, in total, a number of 𝑘. 𝐶𝑚 . 𝐶𝑛𝑣 (𝜇 +
4 6
𝑣)! different words can be formed, namely 15. 𝐶10
𝐶11 . 10!
4 6
Answer: 15. 𝐶10 𝐶11 . 10!
14. In a grocery store there are three types of candy. The candy
is wrapped in three different boxes, each brand in its box. In how many
ways can a set of five boxes be ordered?
Solution (see Problem 11).
Answer: 21.
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Algebraic problems and exercises for high school
15. In order to form the 10 person guard of honor, officers of the
following troops are invited: infantry, aviation, frontier guards, artillery;
navy officers and rocket officers.
In how many ways can the guard of honor be assembled?
Solution. We have 6 categories of officers. Repeating the
reasoning from problem 10, we have to calculate combinations with
repetitions of 6 elements taken 10 times each, as follows
Answer: 3 003.
16. On a shelf there are 𝑚 + 𝑛 different books. Among them 𝑚
have a blue cover, and 𝑛 a yellow cover. The books are permutated in
every way possible. How many positions do the books have, if:
a) the books with the blue cover occupy the first 𝑚 places;
b) the books in the yellow covers sit beside them?
Solution. a) The books with blue covers can be placed in the first
𝑚 places in 𝑃𝑚 = 𝑚! ways. With each such allocation, the books with
yellow covers can be placed in 𝑃𝑛 = 𝑛! ways. Using the multiplication
rule, we have in total 𝑚! · 𝑛! positions in which the blue covers books
occupy the first 𝑚 places.
b) Let the books in blue covers sit beside. It follows that right
beside them on the shelf there can be either 𝑛 books with yellow
covers or 𝑛 − 1, or 𝑛 − 2, ... , or no book with yellow covers. We can
thus place the books with blue covers so that they follow each other in
𝑛 + 1 ways. In each of these positions, the books with yellow covers
can be permutated in any way, also the books with the blue covers can
be permutated in any way. As a result, we will have 𝑚! . 𝑛! . (𝑛 + 1)
different positions for the books.
Answer: a) 𝑚! · 𝑛!; b) 𝑚! · 𝑛! · (𝑛 + 1).
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17. Determine the fourth term of the development of Newton’s
binomial:
Solution. According to the formula (9), the rank 4 term has the
form
Answer: −1792 · 𝑥 17/2.
18. Determine the biggest coefficient in the development of the
binomial
Solution.
If 𝑚 is an even number, i.e. 𝑚 = 2𝑠 , 𝑠 ∈ ℕ∗ , then the
development of the binomial contains 2𝑠 + 1 terms, and according
𝑠
to the 𝑉𝐼𝐼 property, the coefficient 𝐶2𝑠
is the biggest.
If 𝑚 is an uneven number, i.e. 𝑚 = 2𝑠 + 1, 𝑠 ∈ ℕ, according
to the same 𝑉𝐼𝐼 property, the development of the binomial contains
𝑠
𝑠+1
two terms have the biggest coefficients 𝐶2𝑠+1
, 𝐶2𝑠+1
.
𝑠
𝑠
𝑠+1
Answer: 𝐶2𝑠 , if 𝑚 is an even number; 𝐶2𝑠+1 , 𝐶2𝑠+1
if 𝑚 is an
uneven number.
19. Determine which term do not contain 𝑥 in the development
of the binomial:
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Algebraic problems and exercises for high school
Solution.
The term of rank 𝑘 + 1 in the development of this binomial has
the form
This term does not contain 𝑥 only if 𝑘 − 𝑛 = 0 ⇔ 𝑘 = 𝑛. So
the term that doesn’t contain 𝑥 is 𝑇𝑛+1 .
Answer: 𝑇𝑛+1 .
20. In the development of the binomial
determine the terms that contains 𝑎 to the
power of three, if the sum of the binomial coefficients that occupy
uneven places in the development of the binomial is equal to 2 048.
Solution. We will first determine the exponent 𝑛. Based on the
𝑉𝐼 th property, the sum of the binomial coefficients is 2𝑛 . Because the
sum of the binomial coefficients that occupy uneven places in the
development of the binomial is equal to 2048, and based on the 𝑉𝐼 𝑏
property, it is equal to the sum of the coefficients that occupy even
places in the respective development, we have
Therefore, the degree of the binomial is 12. The term of rank
𝑘 + 1 takes the form
Considering the requirements of the problem, we have
and
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
Answer: −264𝑎3 𝑏7.
21. For what value of 𝑛 the binomial coefficients of the second,
third and fourth term from the development of the binomial
(𝑥 + 𝑦)𝑛 forms an arithmetic progression?
Solution. Based on the formula (8), we have
and from the conditions of the problem the following relations issues
The conditions of the problem are verified by the value 𝑛 = 7.
Answer: 𝑛 = 7.
22. Prove that the difference of the coefficients 𝑥 𝑘+1 and 𝑥 𝑘 in
the development of the binomial (1 + 𝑥)𝑛+1 is equal to the difference
of the coefficients of 𝑥 𝑘+1 and 𝑥 𝑘−1 I in the development of the
binomial (1 + 𝑥)𝑛 .
Solution. The coefficients of 𝑥 𝑘+1 and 𝑥 𝑘 in the development of
𝑘+1
𝑘
the binomial (1 + 𝑥)𝑛+1 are 𝐶𝑛+1
and 𝐶𝑛+1
respectively. We assess the
difference
In the development of the binomial (1 + 𝑥)𝑛 , the coefficients of
𝑥 𝑘+1 and 𝑥 𝑘−1 are 𝐶𝑛𝑘+1 and 𝐶𝑛𝑘−1 , respectively. We assess the
difference
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Algebraic problems and exercises for high school
As the members from the right in (∗) and (∗∗) are equal, the
equality of the members from the left results, i.e.
which had to be proven.
23. Comparing the coefficients of 𝑥 in both members of the
equality
prove that
Solution.
In the right member of this equality, the coefficient of 𝑥 𝑘 is
and in the development of the binomial (1 + 𝑥)𝑛+𝑚 , the term of rank
𝑘 + 1 has the form
As the polynomials (1 + 𝑥)𝑚 . (1 + 𝑥)𝑛 and (1 + 𝑥)𝑛+𝑚 are
equal and have the same degree, the equality of the coefficients beside
these powers of 𝑥 result and this also concludes the demonstration of
the equality (𝐴).
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
24. Prove the equality
Solution. Let:
We multiply both members of the last equality with (𝑛 + 1).
We obtain
Returning to the initial expression, we have
that had to be proven.
25. Prove that
Solution. We write the development of the binomial (𝑥 − 1)𝑛 :
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Algebraic problems and exercises for high school
We derive both members of the equality (𝐴) according to 𝑥. We
obtain
We place in (∗∗) 𝑥 = 1. Then
that had to be proven.
26. Prove that the equality
is true.
Solution. We notice that the expression
represents the development of the binomial (1 − 10)2𝑛 = 92𝑛 =
(81)𝑛 , that had to be proven.
27. Bring the expression 𝑃1 + 2𝑃2 + ⋯ + 𝑛𝑃𝑛 to a simpler form.
Solution. We will make the necessary transformations by
applying the method of mathematical induction. Let
For:
𝑛 = 1, we have 𝑃1 = 𝐴1 ⇔ 𝐴1 = 1;
𝑛 = 2, we have 𝑃1 + 2𝑝2 = 𝐴2 ⇔ 1 + 2.2! = 𝐴2 ⇔ 5 = 𝐴2 ⇔
3! − 1 = 𝐴2 ⇔ 𝑃3 − 1 = 𝐴2
𝑛 = 3 , we have 𝑃1 + 2𝑝2 + 3𝑃2 = 𝐴3 ⇔ 𝐴2 + 3𝑃3 = 𝐴3 ⇔
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We assume that for 𝑛 = 𝑘 the equality (∗) takes the form
We calculate the value of the expression (∗) for 𝑛 = 𝑘 + 1. We
have
Based on the principle of mathematical induction, we reach the
conclusion that
Answer:
28. Prove that indifferent of what 𝑚, 𝑛 ∈ ℕ, are 𝑚! . 𝑛! divides
(𝑚 + 𝑛)!
Solution. According to the definition 3,
(𝑚+𝑛)!
𝑚! . 𝑛!
𝑛
= 𝐶𝑚+𝑛
is the
number of the subsets that have 𝑛 elements of a set with (𝑚 + 𝑛)
𝑛
elements, namely 𝐶𝑚+𝑛
is a natural number. Consequently,
(𝑚+𝑛)!
𝑚! . 𝑛!
an integer number, or this proves that 𝑚! . 𝑛! divides (𝑚 + 𝑛)!
29. Deduce the equality
Solution. We use property𝑋. We have:
As a result,
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Algebraic problems and exercises for high school
that had to be proven.
30. Calculate the sum
Solution. We notice that term 𝑎𝑛 of this sum can be transformed
as it follows:
Then the sum 𝑆𝑛 takes the form
Answer:
31. Solve the equation
Solution.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
Because 𝐶𝑥3 has meaning only for 𝑥 ≥ 3, it follows that the
solution of the initial equation is 𝑥 = 7.
Answer: 𝑥 = 7.
32. Solve the equation
Solution.
Answer: 𝑥 = 5.
33. Solve the equation
Solution. As:
We have:
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Algebraic problems and exercises for high school
As 𝑦 ∈ ℕ and 𝑦 ≤ 𝑥, we have 𝑦 ∈ {0, 1 ,2, 3, 4, 5, 6, 7, 8}.
Answer: 𝑥 = 8; 𝑦 ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8}.
34. Determine the values of 𝑥 that verify the equality
Solution.
because the solutions of the equation 𝑥 2 + 𝑥 − 5 = 0 are irational
numbers.
Answer: 𝑥 = 3.
35. Solve the equation
Solution.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
Then 𝑛 ∈ {0, 1,2,3,4,5,6, 7,8}.
Answer: 𝑥 = 9, 𝑛 ∈ {0, 1,2,3,4,5,6, 7,8}.
36. Solve the system of equations
Solution. From the conditions of the problem, we have 𝑥, 𝑦 ∈ ℕ
with 𝑥 ≥ 1 and ≤. Based on the formulas (1) − (3), we have
The divisors of the free term in (∗) are the numbers
±1; ±2; ±3; ±4; ±5; ±6; ±7; ±8; ±9; ±10; . . . .
We use Homer’s scheme and the Bezout theorem to select the
numbers that are solutions of the equation (∗). As 𝑦 ∈ ℕ, we will
verify only the natural numbers. It verifies that numbers {1, 2, 3, 4, 5, 6}
are not solutions of the equation (∗).
We verify 𝑦 = 7.
So
because for 𝑦 > 7, the expression 𝑦 4 − 3𝑦 3 + 14𝑦 2 + 48𝑦 + 360 > 0.
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Algebraic problems and exercises for high school
Hence, the solution of the system (𝐵) is the pair (5,7).
Answer: {(5,7)}.
37. Find x and y, if
Solution. We will first bring to a simpler form the expression
As a result, the system (𝐶) takes the form
Answer: {(7, 3)}.
38. Determine the values of 𝑥, so that
Solution. From the enunciation it follows that 𝑥 ∈ ℕ. Then
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
and 𝑥 ∈ ℕ. So 𝑥 ∈ {0, 1, 2, 3, 4, 5, 6}.
Answer: 𝑥 ∈ {0, 1, 2, 3, 4, 5, 6}.
39. Determine the values of 𝑥 that verify the inequation
Solution. (∗) has meaning for 𝑥 ∈ ℕ and 𝑥 ≥ 3. Because
it follows that
Answer: 𝑥 ∈ {3, 4, 5, 6}.
40. Solve the system of inequations:
Solution.
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Algebraic problems and exercises for high school
Answer: 𝑥 ∈ {2,3}.
3.3. Suggested exercises
1. A commission is formed of one president, his assistant and
five other persons. In how many ways can the members of the
commission distribute the functions among themselves?
2. In how many ways can three persons be chosen from a group
of 20 persons to accomplish an assignment?
3. In a vase there are 10 red daffodils and 4 pink ones. In how
many ways can three flowers be chosen from the vase?
4. The padlock can be unlocked only if a three digit number is
correctly introduced out of five possible digits. The number is guessed,
randomly picking 3 digits. The last guess proved to be the only
successful one. How many tries have proceeded success?
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
5. On a shelf there are 30 volumes. In how many ways can the
books be arranged, so that volumes 1 and 2 do not sit beside each
other on the shelf?
6. Four sharpshooters have to hit eight targets (two targets
each). In how many ways can the targets be distributed among the
sharpshooters?
7. How many four digit numbers, made out of digits 0, 1, 2, 3, 4,
5, contain digit 3, if: a) the digits do not repeat in the number; b) the
digits can repeat themselves?
8. In the piano sections there are 10 participating persons, in the
reciter section, 15 persons, in the canto section, 12 persons, and in the
photography section - 20 persons. In how many ways can there a team
be assembled that contains 4 reciters, 3 pianists, 5 singers and a
photographer?
9. Seven apples and three oranges have to be placed in two
bags such that every bag contains at least one orange and the number
of fruits in the bags is the same. In how many ways can this distribution
be made?
10. The matriculation number of a trailer is composed of two
letters and four digits. How many matriculation numbers can there be
formed using 30 letters and 10 digits?
11. On a triangle side there are taken 𝑛 points, on the second
side there are taken 𝑚 points, and on the third side, 𝑘 points.
Moreover, none of the considered points is the triangle top. How many
triangles with tops in these points are there?
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Algebraic problems and exercises for high school
12. Five gentlemen and five ladies have to be sited around a
table so that no two ladies and no two gentlemen sit beside each other.
In how many ways can this be done?
13. Two different mathematic test papers have to be distributed
to 12 students. In how many ways can the students be arranged in two
rows so that the students that sit side by side have different tests and
those that sit behind one another have the same test?
14. Seven different objects have to be distributed to three
persons. In how many ways can this distribution be made, if one or two
persons can receive no object?
15. How many six digit numbers can be formed with the digits 1,
2, 3, 4, 5, 6, 7, in such a way that the digits do not repeat, and that the
digits at the beginning and at the end of the number are even?
16. How many different four digit numbers can be formed with
the digits 1, 2, 3, 4, 5, 6, 7, 8, if in each one the digit one appears only
once, and the other digits can appear several times?
17. To award the winners of the Mathematics Olympiad, three
copies of a book, two copies of another book and one copy of a third
book have been provided. How many prizes can be awarded if 20
persons have attended the Olympiad, and no one will receive two
books simultaneously? Same problem, if no one will receive two copies
of the same book but can receive two or three different books?
18. The letters of the Morse alphabet is comprised of symbols
(dots and dashes). How many letters can be drawn, if it is required that
each letter contains no more than five symbols?
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
19. To find their lost friend, some tourists have formed two
equal groups. Only four persons among them know the surroundings.
In how many ways can the tourists divide so that each group has two
persons that know the surroundings and considering they are 16
persons in total?
20. Each of the 10 radio operators located at point A is trying to
contact each of the 20 radio operators located at point B. How many
different options for making contact are there?
Prove the equalities:
(where 𝑛, 𝑘 ∈ ℕ, 𝑛 > 𝑘 + 3);
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Algebraic problems and exercises for high school
Solve the equations and the systems of equations:
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
Solve the inequations and the systems of inequations:
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Algebraic problems and exercises for high school
107. Determine the fifth term from the development of the
binomial
108. Determine the middle term from the development of the
binomial
109. Determine the value of exponent 𝑚 from the development
of the binomial (1 + 𝑎)𝑚 , if the coefficient of the 5th term is equal to
the coefficient of the 9th term.
110. Determine 𝐴2𝑛 , if the 5th term from the development of the
binomial
doesn’t depend on 𝑥.
111. In the development of the binomial
the coefficient of the third term is equal to 28. Determine the middle
term from the development of the binomial.
112. Determine the smallest value of 𝑚′𝑠 exponent from the
development of the binomial (1 + 𝑎)𝑚 , if the relation of the
coefficients of two neighboring arbitrary is equal to 7: 15.
113. Determine the term from the development of the binomial
that contains 𝑥 3.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
114. Determine the term from the development of the binomial
that doesn’t depend on 𝑎.
115. Determine the term from the development of the binomial
that doesn’t contain 𝑎, if the sum of
the coefficients of the first three terms from the development is equal
to 79.
116. Determine the term from the development of the binomial
that doesn’t contain 𝑎.
117. Determine the free term from the development of the
binomial
118. Determine the third term from the development of the
binomial
2048.
if the sum of the binomial coefficients is
119. Determine the term from the development of the binomial
that contains 𝑏6 , if the relation of the
binomial coefficients of the fourth and second term is equal to 187.
120. Determine the term from the development of the binomial
that doesn’t contain 𝑥, if the sum of the
coefficients of the second term from the beginning of the
development and the third term from the end of the development is
equal to 78.
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Algebraic problems and exercises for high school
121. The relation of the coefficient of the third term to the
coefficient of the 5th term in the development of the binomial
is equal to 2/7. . Determine the term from the
development of the binomial that contains 𝑥 −5/2.
122. Determine 𝑥, 𝑦 and 𝑧, if it is known that the 2nd, 3rd and 4th
terms from the development of the binomial
240, 720, 1080, respectively.
are equal to
123. For what value of the exponent 𝑛, the coefficients of the
2nd, the 3rd and the 4th term from the development of the binomial
form an arithmetic progression?
124. Determine the terms from the development of the
binomial
that do not contain irrationalities.
125. How many rational terms does the following development
of the binomial contain
126. Determine the ranks of three consecutive terms from the
development of the binomial (𝑎 + 𝑏)23 the coefficients of which form
an arithmetic progression.
127. Determine the term from the development of the binomial
4
(√𝑥 + √𝑥 −3)𝑛 that contains 𝑥 6.5 , if the 9th term has the biggest
coefficient.
128. The 3rd term from the development of the binomial
1
(𝑠𝑥 + 𝑥 2 )𝑚 doesn’t contain 𝑥. For which value of 𝑥 is this term equal to
the 2nd term from the development of the binomial (1 + 𝑥 3 )30 ?
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
129. For what positive values of 𝑥 the biggest term from the
development of the binomial (5 + 3𝑥)10 is the 4th term?
130. In the development of the binomial (√𝑥 +
1
4
2 √𝑥
)𝑛 the first
three terms form an arithmetic progression. Determine all rational
terms from the development of this binomial.
131. Determine the values of 𝑥 for which the difference between
the 4th and the 6th term from the development of the binomial
√2𝑥
16
( 16 +
√8
√32 𝑚
√2𝑥
) is equal to 56, if it is known that the exponent 𝑚 of the
binomial is smaller by 20 than the binomial coefficient of the 3rd term
from the development of this binomial.
132. Given that 𝑛 is the biggest natural number only if
𝑙𝑜𝑔𝑙/3 𝑛 + 𝑙𝑙𝑜𝑔𝑛/3 𝑛 > 0, determine the term that contains 𝑏2 from
3
the development of the binomial (√𝑎 − √𝑏)𝑛 .
133. Determine 𝑥 for which the sum of the 3rd and al 5th term
from the development of the binomial (√2𝑥 + √21−𝑥 )𝑛 is equal to
135, knowing that the sum of the last three binomial coefficients is 22.
134. Determine 𝑥 , knowing that the 6th term from the
𝑥
development of the binomial (𝑎 + 𝑏)𝑛 , where 𝑎 = √2lg(10−3 ) , 𝑏 =
5
√2(𝑥−2)𝑙𝑔3 is 21, and the coefficients of the binomial of the terms
ranked 2, 3 and 4 are respectively the 1st, the 3rd and the 5th term of an
arithmetic progression.
135. Are there any terms independent of 𝑥 in the development
of the binomial
Write these terms.
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Algebraic problems and exercises for high school
136. How many terms from the development of the binomial
5
3
( √3 + √7)36 are integer terms?
137. In the development of the binomial
the first three coefficients form an arithmetic
progression. Determine all terms from the development of the
binomial that contain the powers of 𝑦 with a natural exponent.
138. Determine 𝑥, if the 3rd term from the development of the
binomial
is equal to 106.
139. In the development of (1 + 𝑥 − 𝑥 2 )25 find the term
corresponding to the exponent of 𝑥 that is 3 times as big as the sum of
all the development’s coefficients sum.
140. Determine the rank of the biggest term from the
development of the binomial (𝑝 + 𝑞)𝑛 according to the descending
powers of 𝑝, assuming that 𝑝 > 0; 𝑞 > 0; 𝑝 + 𝑞 = 1. In which
conditions:
a) the biggest term will be the first?
b) the biggest term will be the last?
c) the development will contain two consecutive equal terms
that are bigger than all the other terms of the development?
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
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Algebraic problems and exercises for high school
Answers
Note. For chapters 1 and 2 only the less „bulky” answers have
been provided and those that are relatively more complicated (in our
own opinion).
1. Sets. Operations with sets
(because 𝐴 consists of even natural numbers).
𝑏
a)if 𝑎𝑑 − 𝑏𝑐 = 0 ⇒ 𝐶 = { } ; b) if 𝑎𝑑 − 𝑏𝑐 ≠ 0 ⇒ 𝐶 ℎ𝑎𝑠 𝑝 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠
𝑑
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
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Algebraic problems and exercises for high school
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
2. Relations, functions
4. 1) 𝑥 + 𝑦 = 6; 2) 𝑦 = 𝑥 + 1; 3) 𝑥 < 𝑦; 4) max (𝑥, 𝑦) > 4; 5) min
(𝑥, 𝑦) = 2; 6) cmmdc (𝑥, 𝑦) = 2; 7) 𝑥 is even or 𝑦 = 6; 8) 𝑥 = 𝑦 2.
5. 1) transitive; 2) symmetrical; 3) symmetrical and transitive; 4)
reflexive, transitive; 5) reflexive; 6) reflexive, symmetrical; 7), 8)
reflexive, symmetrical, transitive; 9) reflexive, transitive; 10) transitive
etc.
6. 1) 𝛿𝛼 = 𝜌𝛼 = ℕ, symmetrical; 2) 𝛿𝛼 = 𝜌𝛼 = ℕ, reflexive; 3) 𝛿𝛼 =
𝜌𝛼 = ℕ, symmetrical, antireflexive; 4) 𝛿𝛼 = {1, 4, 9, … , 𝑛2 , … },𝜌𝛼 = ℕ
anti-symmetrical; 5) ) 𝛿𝛼 = 𝜌𝛼 = ℕ,, reflexive, symmetrical, transitive
etc. 6) 𝛿𝛼 = 𝜌𝛼 = {1, 2, 3, 5, 6, 10, 15, 30}, antireflexive, symmetrical;
7) 𝛿𝛼 = ℕ, 𝜌𝛼 = {3, 4, 5, . . . } , antireflexive, symmetrical; 8) 𝛿𝛼 =
𝜌𝛼 = ℕ , reflexive, anti-symmetrical, transitive; 9) 𝛿𝛼 = ℕ, 𝜌𝛼 =
{3,4,5, . . . } , antireflexive, anti-symmetrical, transitive; 10) 𝛿𝛼 =
ℕ, 𝜌𝛼 = {0, 2, 4, . . . }, anti-symmetrical; 11) 𝛿𝛼 = 𝜌𝛼 = ℕ , reflexive,
symmetrical, transitive; 12) 𝛿𝛼 = 𝜌𝛼 = ℕ, symmetrical.
𝑒
9. 1) â = {𝑎, 𝑒/𝑎}, 𝑎 ∈ ℝ, 𝑎 ≠ ; 2) 𝑎 = √𝑒, we have â = {√𝑒}.
𝑎
10. â = {𝑥 ∈ ℝ│𝑥 = 𝑎 + 2𝑘𝜋 𝑜𝑟 𝑥 = 𝜋 − 𝑎 + 2𝑚𝜋, 𝑘, 𝑚 ∈ ℤ}.
11. a) yes;
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Algebraic problems and exercises for high school
23. 1) bijective; 2) not bijective; 3) bijective; 4) not bijective; 5) bijective:
6) not bijective; 7) bijective; 8) not bijective; 9) injective; 10) not
injective; 11) injective; 12) surjective; 13) not surjective; 14) surjective;
15) surjective.
32. 1) yes; 2) no; 3) yes; 4) yes; 5) yes; 6) no; 7) no; 8) no.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
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Algebraic problems and exercises for high school
3. Elements of combinatorics
12. 110. 240. 111. 70(1 − 𝑥 2 )2. 112. 𝑚 =
22𝑘+15
7
, the smallest value
6
6
5
𝑘 = 6, then 𝑚 = 21. 𝐶16
. 𝑥 3. 114. 𝑇6 = 𝐶15
. 115. 𝑇5 = 𝐶12
. 6−7. 116.
137. 𝑛 = 8, we have 𝑇1 = 𝑦 4 , 𝑇5 =
35
8
𝑦; 𝑛 = 4, we have 𝑇1 = 𝑦 2 .
138. 10.
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Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache
References
1. Goian I., Grigor R., Marin V.: Baccalaureate "Mathematics" 1993 1995. Chisinau: Amato, 1996.
2. Goian I., Grigor R., Marin V.: Baccalaureate "Mathematics" 1996.
Chisinau: Evrica, 1997.
3. Ionescu-Țău C., Muşat Şt.: Exercises and mathematics problems for
grades IX - X. Bucharest: Didactică and pedagogică, 1978.
4. Militaru C.: Algebra. Exercises and problems for high school and
university admission. Bucharest: Alux, 1992.
5. Năstăsescu C., Niţă C., Popa S.: Mathematics. Algebra. Textbook for
grade X-a. Bucharest: Didactică and pedagogică, 1994.
6. Petrică I., Lazăr I.: Math tests for rank I-a and a II-a high school.
Bucharest: Albatros, 1981.
7. Sacter O.: Math problems Workbook suggested at maturity exams and
at admissions in institutions and universities. Bucharest: Tehnică,
1963.
8. Stamate I., Stoian I.: Algebra problems workbook for high schools.
Bucharest: Editura didactică and pedagogică, 1971.
9. Şahno C.I.: Elementary mathematics problems workbook. Translation
from Russian by I. Cibotaru. Chisinau: Lumina, 1968.
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Algebraic problems and exercises for high school
In this book, you will find algebra exercises and problems, grouped by
chapters, intended for higher grades in high schools or middle schools
of general education. Its purpose is to facilitate training in mathematics
for students in all high school categories, but can be equally helpful in a
standalone workout. The book can also be used as an extracurricular
source, as the reader shall find enclosed important theorems and
formulas, standard definitions and notions that are not always included
in school textbooks.
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